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Non-concave utility maximisation on the positive real axis in discrete time

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Abstract

We treat a discrete-time asset allocation problem in an arbitrage-free, generically incomplete financial market, where the investor has a possibly non-concave utility function and wealth is restricted to remain non-negative. Under easily verifiable conditions, we establish the existence of optimal portfolios.

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Notes

  1. Here \(x^{+}\triangleq \max \left\{ x,0\right\} \) and \(x^{-}\triangleq -\min \left\{ x,0\right\} \) for every \(x\in \mathbb {R}\). Furthermore, in order to make the notation less heavy, given any function \(f:~X\rightarrow \mathbb {R}\), we shall write henceforth \(f^{\pm }\left( x\right) \triangleq \left[ f\left( x\right) \right] ^{\pm }\) for all \(x\in X\).

  2. To be precise, in the cited lemma there is strict inequality in (2.5) and it is required to hold for \(\lambda >1\) only. As easily seen, it works also for our version.

  3. Given a set \(E\subseteq X\times Y\), we recall that the projection of \(E\) on \(X\) is

    $$\begin{aligned} {\text {Proj}}_{X}\!\left( E\right) \triangleq \left\{ x\in X\text {: }\exists \,y\in Y\text { such that }\left( x,y\right) \in E\right\} . \end{aligned}$$

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Acknowledgments

L. Carassus thanks LPMA (UMR7599) for support. A. M. Rodrigues gratefully acknowledges the financial support of FCT–Fundação para a Ciência e Tecnologia (Portuguese Foundation for Science and Technology) through the Doctoral Grant SFRH/BD/69360/2010. Part of this research was carried out while M. Rásonyi. and A. M. Rodrigues were affiliated with the School of Mathematics, University of Edinburgh, Scotland, U.K.

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Correspondence to Miklós Rásonyi.

Appendix

Appendix

We are staying in the setting of Sect. 4.

Example 5.1

It is not difficult to find (non-random) utilities which are bounded above and yet have non-zero (actually, infinite) asymptotic elasticity, as the example below shows. The construction below is inspired by the proof of Lemma 6.5 in Kramkov and Schachermayer [15]. Let \(f:[0,\infty )\rightarrow \mathbb {R}\) be the function which takes the values

$$\begin{aligned}&f\left( n\right) \triangleq \frac{1}{2}-\frac{1}{n+1}=\frac{n-1}{2\left( n+1\right) },\\&f\left( n+1/2-a_{n}\right) \triangleq f\left( n\right) +a_{n},\\&f\left( n+1/2+a_{n}\right) \triangleq f\left( n+1\right) -a_{n}, \end{aligned}$$

with \(a_{n}\triangleq 1/\left[ 4\left( n+1\right) \left( n+2\right) \right] \) for every \(n\in \mathbb {N}\), and which is linear between the points where it has been defined.

Clearly \(f\left( +\infty \right) =1/2\) and \(f\left( 1\right) =0\). We also note that \(f\left( 0\right) =-1/2>-\infty \). Moreover, the piecewise linearity of \(f\) and trivial computations yield

$$\begin{aligned} f'\left( x\right) =\frac{f\left( n+1/2+a_{n}\right) -f\left( n+1/2-a_{n}\right) }{2a_{n}}=1 \end{aligned}$$

for any \(x\in \left( n+1/2-a_{n},n+1/2+a_{n}\right) \), so in particular \(f'\left( n+1/2\right) \) equals \(1\). Furthermore, we have the following inequality,

$$\begin{aligned} \frac{f\left( n+1/2\right) }{n+1/2}\le \frac{f\left( n+1\right) }{n+1/2}=\frac{n}{\left( n+2\right) \left( 2n+1\right) }, \end{aligned}$$

thus combining all of the above gives \(\lim _{n\rightarrow +\infty }\left( n+1/2\right) f'\left( n+1/2\right) /f\left( n+1/2\right) =+\infty \), and hence \(AE_{+}\left( u\right) =+\infty \) in the classical sense of Kramkov and Schachermayer [15]. We finish by noticing that, as in the proof of Lemma 6.5 in Kramkov and Schachermayer [15], \(f\) can be slightly modified in such a way that it becomes smooth and our conclusion is still valid. So, as mentioned in Remark 2.11, Lemma 6.3 of Kramkov and Schachermayer [15] applies and

$$\begin{aligned} \limsup _{x\rightarrow +\infty }\frac{x\,f'\left( x\right) }{f\left( x\right) }= \inf \left\{ \gamma >0\,\text {:}\,\exists \,\overline{x}\ge 0 \text {s.t. a.e.}\, f\left( \lambda x,\cdot \right) \le \lambda ^{\gamma }f\left( x,\cdot \right) , \ \forall \,\lambda \ge 1,\ \forall \,x\ge \overline{x}\right\} , \end{aligned}$$

with the usual convention that the infimum of the empty set is \(+\infty \), and \(f\) has an infinite asymptotic elasticity in both senses. Nevertheless, choosing \(\overline{x}=1\), we have that \(0 \le f(x) \le 1/2\) for \(x \ge \overline{x}\), and Assumption 2.10 holds true for \(u(x)=f(x)\).

Proof of Lemma 2.12

Consider an arbitrary \(x\in [0,\overline{x})\). Then we can use the fact that \(u\) is non-decreasing and inequality (2.4) to obtain for a.e. \(\omega \in \Omega \) that

$$\begin{aligned} u\left( \lambda x,\omega \right) \le u\left( \lambda \overline{x},\omega \right) \le \lambda ^{\overline{\gamma }}u\left( \overline{x},\omega \right) +\lambda ^{\overline{\gamma }}c\left( \omega \right) \le \lambda ^{\overline{\gamma }}u^{+}\left( \overline{x},\omega \right) +\lambda ^{\overline{\gamma }}c\left( \omega \right) \end{aligned}$$

for any \(\lambda \ge 1\). As \(c \ge 0\) a.s. this implies that \(u^{+}\left( \lambda x,\omega \right) \le \lambda ^{\overline{\gamma }}u^{+}\left( \overline{x},\omega \right) +\lambda ^{\overline{\gamma }}c\left( \omega \right) .\)

On the other hand, for a.e. \(\omega \in \Omega \) and every \(x\ge \overline{x}\), we have by (2.4) that

$$\begin{aligned} u\left( \lambda x,\omega \right) \le \lambda ^{\overline{\gamma }}\left[ u\left( x,\omega \right) +c\left( \omega \right) \right] \le \lambda ^{\overline{\gamma }}\left[ u^{+}\left( x,\omega \right) +c\left( \omega \right) \right] \end{aligned}$$

for all \(\lambda \ge 1\). Using again that \(c \ge 0\) a.s. we get that \(u^{+}\left( \lambda x,\omega \right) \le \lambda ^{\overline{\gamma }}\left[ u^{+}\left( x,\omega \right) +c\left( \omega \right) \right] .\) Hence, choosing \(C\left( \omega \right) \triangleq u^+\left( \overline{x},\omega \right) +c\left( \omega \right) \ge 0\) and combining the previous inequalities we get for a.e. \(\omega \in \Omega \)

$$\begin{aligned} u^{+}\left( \lambda x,\omega \right)&\le \max \left\{ \lambda ^{\overline{\gamma }}\left[ u^{+}\left( \overline{x},\omega \right) +\!c\!\left( \omega \right) \right] ,\lambda ^{\overline{\gamma }}\left[ u^{+}\left( x,\omega \right) +c\left( \omega \right) \right] \right\} \le \! \lambda ^{\overline{\gamma }}u^{+}\left( x,\omega \right) +\lambda ^{\overline{\gamma }}C\left( \omega \right) \end{aligned}$$

for all \(\lambda \ge 1\), \(x\ge 0\). Lastly, note that \(\mathbb {E}_{\mathbb {P}}\left[ C\right] <+\infty \) since \(\mathbb {E}_{\mathbb {P}}\left[ u^+\left( \overline{x},\cdot \right) \right] <+\infty \) and \(\mathbb {E}_{\mathbb {P}}\left[ c\right] <+\infty \) by Assumption 2.10.\(\square \)

Proof of Lemma 4.8

Let \(\Theta \) denote the set of functions from \(\{1,\ldots ,d\}\) to \(\{-\sqrt{d},\sqrt{d}\}\), and let \(x>0\). Then we have by Lemma 4.7 that, for all \(\xi \in \widetilde{\Xi }^d(x)\),

$$\begin{aligned} x+\left\langle \xi ,Y\right\rangle \le x\left( 1+\left\| Y\right\| /\beta \right) \le x\left[ 1+\left( 1/\beta \right) \max _{\tau \in \Theta }\left\langle \tau ,Y\right\rangle \right] \,\text {a.s.}, \end{aligned}$$

hence

$$\begin{aligned} V^{+}\left( x+\left\langle \xi \left( \cdot \right) ,Y\left( \cdot \right) \right\rangle ,\cdot \right)&\le V^{+}\left( x\left[ 1+\left( 1/\beta \left( \cdot \right) \right) \max _{\tau \in \Theta }\left\langle \tau ,Y\left( \cdot \right) \right\rangle \right] ,\cdot \right) \\&\le \sum _{\tau \in \Theta }V^{+}\left( x\left[ 1+\left\langle \tau /\beta \left( \cdot \right) ,Y\left( \cdot \right) \right\rangle \right] ,\cdot \right) \,\text {a.s.,} \end{aligned}$$

where we define \(V^{+}\left( x,\omega \right) \triangleq 0\) for \(x<0\).

We know from Proposition 4.2 of Rásonyi and Stettner [19] that there is a random set \(M\left( 1\right) \in \fancyscript{G}\otimes \fancyscript{B}\left( \mathbb {R}^d\right) \) such that \(\xi \in M\left( 1\right) \) a.s. if and only if \(\xi \in \tilde{\Xi }^{d}\left( 1\right) \). Denoting the linear span of \(M\left( 1\right) \left( \omega \right) \) by \(R\left( 1\right) \left( \omega \right) \), \(R\) is again in \(\fancyscript{G}\otimes \fancyscript{B}\left( \mathbb {R}^d\right) \) by Proposition 4.3 of Rásonyi and Stettner [19]. It suffices to prove our lemma separately on the sets \(\left\{ \omega \in \Omega \,\text {: }\text {dim}\,R\left( 1\right) \left( \omega \right) =k\right\} \), with \(k\in \left\{ 0,\ldots ,d\right\} \), since these sets are in \(\fancyscript{G}\) (see the proof of Proposition 4.3 in Rásonyi and Stettner [19]). Applying verbatim the arguments of Lemma 2.3 in Rásonyi and Stettner [19], one can show the existence of \(g\in \Xi ^{d}\left( 1\right) \) and of \(\varepsilon _{\tau }\in \Xi ^{1}\) with \(\varepsilon _{\tau }\in (0,1)\) such that \(\widetilde{g}_{\tau }\triangleq g+\varepsilon _{\tau }\left( {\tau }/\beta -g\right) \) belongs to \(M(1)\) and thus to \(\Xi ^{d}\left( 1\right) \) (with the notation of the cited lemma, \(\fancyscript{G}=\fancyscript{H}\), \(g=\rho \), \(K\theta _i=\tau /\beta \), \(\varepsilon _{\tau }=\psi _i\) and \(i=\tau \)). It follows that, for \(x\le 1\),

$$\begin{aligned} V^{+}\left( x+\left\langle \xi \left( \cdot \right) ,Y\left( \cdot \right) \right\rangle ,\cdot \right) \le \sum _{\tau \in \Theta }V^{+}\left( 1+\left\langle \tau /\beta \left( \cdot \right) ,Y\left( \cdot \right) \right\rangle ,\cdot \right) \,\text {a.s.}, \end{aligned}$$

and almost surely for \(x>1\),

$$\begin{aligned} V^{+}\left( x+\left\langle \xi \left( \cdot \right) ,Y\left( \cdot \right) \right\rangle ,\cdot \right) \le \sum _{\tau \in \Theta }x^{\overline{\gamma }}\left[ V^{+}\left( 1+\left\langle \tau /\beta \left( \cdot \right) ,Y\left( \cdot \right) \right\rangle ,\cdot \right) + \overline{C}\left( \cdot \right) \right] \end{aligned}$$

by Assumption 4.6. Note that if \(\tau \) is such that \(\{1+\left\langle \tau /\beta \left( \cdot \right) ,Y\left( \cdot \right) \right\rangle < 0\}\) has strictly positive probability then on this set the preceding inequality holds true since \(\overline{C} \ge 0\) a.s.

Applying the same assumption again (and with the same remark), we get a.s. that

$$\begin{aligned}&V^{+}\left( 1+\left\langle \frac{{\tau }}{\beta \left( \cdot \right) },Y\left( \cdot \right) \right\rangle ,\cdot \right) \\&\qquad \le \frac{1}{{\varepsilon _{\tau }\left( \cdot \right) }^{\overline{\gamma }}}\left[ V^{+}\left( \varepsilon _{\tau }\left( \cdot \right) \left[ 1+\left\langle \frac{{\tau }}{\beta \left( \cdot \right) },Y\left( \cdot \right) \right\rangle \right] ,\cdot \right) +\overline{C}\left( \cdot \right) \right] \\&\qquad \le \frac{1}{{\varepsilon _{\tau }\left( \cdot \right) }^{\overline{\gamma }}}\left[ V^{+}\left( \varepsilon _{\tau }\left( \cdot \right) \big [1+\left\langle G\left( \cdot \right) ,Y\left( \cdot \right) \right\rangle \big ]+\varepsilon _{\tau }\left( \cdot \right) \left\langle \frac{\tau }{\beta \left( \cdot \right) }-G\left( \cdot \right) ,Y\left( \cdot \right) \right\rangle ,\cdot \right) +\overline{C}\left( \cdot \right) \right] \\&\qquad \le \frac{1}{{\varepsilon _{\tau }\left( \cdot \right) }^{\overline{\gamma }}}\left[ V^{+}\left( 1+\left\langle G\left( \cdot \right) ,Y\left( \cdot \right) \right\rangle +\varepsilon _{\tau }\left( \cdot \right) \left\langle \frac{\tau }{\beta \left( \cdot \right) }-G\left( \cdot \right) ,Y\left( \cdot \right) \right\rangle ,\cdot \right) +\overline{C}\left( \cdot \right) \right] \\&\quad =\frac{1}{{\varepsilon _{\tau }\left( \cdot \right) }^{\overline{\gamma }}}\left[ V^{+}\left( 1+\left\langle \widetilde{g}_{\tau }\left( \cdot \right) ,Y\left( \cdot \right) \right\rangle ,\cdot \right) +\overline{C}\left( \cdot \right) \right] , \end{aligned}$$

where the last inequality holds true since \(1+\left\langle g,Y\right\rangle \ge 0\) a.s. Now let

$$\begin{aligned} L'\left( \cdot \right) \triangleq \sum _{\tau \in \Theta }\left( \frac{1}{{\varepsilon _{\tau }\left( \cdot \right) }^{\overline{\gamma }}}\left[ V^{+}\left( 1+\left\langle \widetilde{g}_{\tau }\left( \cdot \right) ,Y\left( \cdot \right) \right\rangle ,\cdot \right) +\overline{C}\left( \cdot \right) \right] +\overline{C}\left( \cdot \right) \right) +V^{+}\left( 0,\cdot \right) , \end{aligned}$$

where the last term is added to cover the case \(x=0\) as well. By assumption, \(\mathbb {E}_{\mathbb {P}}\left[ \left. \overline{C}\right| \fancyscript{G}\right] <+\infty \) a.s., and by Assumption 4.4 we have \(\mathbb {E}_{\mathbb {P}}\left[ \left. V^{+}\left( 1+\left\langle \widetilde{g}_{\tau }\left( \cdot \right) ,Y\left( \cdot \right) \right\rangle , \cdot \right) \right| \fancyscript{G}\right] <+\infty \) and \(\mathbb {E}_{\mathbb {P}}\left[ \left. V^{+}\left( 0, \cdot \right) \right| \fancyscript{G}\right] <+\infty \) a.s., so the proof is complete since \(\varepsilon _{\tau }\) is \(\fancyscript{G}\)-measurable.\(\square \)

Proof of Lemma 4.9

Let us first choose, for each positive rational number \(q\), a version \(F\left( q,\omega \right) \) of \(\mathop {\mathrm{ess\,sup\,}}\limits _{\xi \in \Xi ^{d}\left( q\right) }\mathbb {E}_{\mathbb {P}}\left[ \left. V\left( q+\left\langle \xi \left( \cdot \right) ,Y\left( \cdot \right) \right\rangle ,\cdot \right) \right| \fancyscript{G}\right] \).

Next, for any pair \(q_{1}<q_{2}\) of positive rational numbers, and any \(\xi \in \Xi ^{d}\left( q_{1}\right) \), one has \(\xi \in \Xi ^{d}\left( q_{2}\right) \), as obviously \(q_{1}+\left\langle \xi ,Y\right\rangle <q_{2}+\left\langle \xi ,Y\right\rangle \). So we have by Assumption 4.3 that

$$\begin{aligned} V\left( q_{1}+\left\langle \xi \left( \omega \right) ,Y\left( \omega \right) \right\rangle ,\omega \right) \le V\left( q_{2}+\left\langle \xi \left( \omega \right) ,Y\left( \omega \right) \right\rangle ,\omega \right) \end{aligned}$$

for \(\mathbb {P}\)-a.e. \(\omega \in \Omega \). We then conclude from the monotonicity of the conditional expectation and taking essential supremum that \(F\left( q_{1},\cdot \right) \le F\left( q_{2},\cdot \right) \) a.s. Similarly, for every positive \(q\in \mathbb {Q}\), \(F\left( q,\omega \right) <+\infty \) from Assumption 4.4. Thus one can find a set \(A \in \fancyscript{G}\) of full probability such that, for every \(\omega \in A\), the mapping \(q \mapsto F\left( q,\omega \right) \) is non-decreasing and finite-valued on the set of positive rational numbers.

So let us specify, for each \(x\in \left[ \left. 0,+\infty \right) \right. \) and \(\omega \in A\),

$$\begin{aligned} G\left( x,\omega \right) \triangleq \inf _{\begin{array}{c} q\in \mathbb {Q}\\ q> x \end{array}} F\left( q,\omega \right) , \end{aligned}$$

and define \(G\left( x,\omega \right) =0\) if \(\omega \in \Omega \setminus A\). Clearly, when \(x\ge 0\) is in \(\mathbb {Q}\) then \(G\left( x,\cdot \right) \ge F\left( x,\cdot \right) \) a.s. In addition, for each \(x\ge 0\) we have that \(G\left( x,\cdot \right) \) is \(\fancyscript{G}\)-measurable. We shall split the remainder of the proof into five separate parts.

(i) :

With the above definition, it is straightforward to check that, for each \(\omega \in \Omega \), the function \(G\left( \cdot ,\omega \right) \) is non-decreasing. It is also clear that \(G\left( x,\omega \right) <+\infty \) for all \(x\ge 0\) and \(\omega \in \Omega \).

(ii) :

We proceed to show that, for all \(x\in \left[ \left. 0,+\infty \right) \right. \),

$$\begin{aligned} G\left( x,\cdot \right) =\mathop {\mathrm{ess\,sup\,}}\limits _{\xi \in \Xi ^{d}\left( x\right) }\mathbb {E}_{\mathbb {P}}\left[ \left. V\left( x+\left\langle \xi \left( \cdot \right) ,Y\left( \cdot \right) \right\rangle ,\cdot \right) \right| \fancyscript{G}\right] \,\text {a.s.} \end{aligned}$$

In order to do so, let us fix an arbitrary \(x\in \left[ \left. 0,+\infty \right) \right. \). Then, for every \(q\in \mathbb {Q}\), \(q>x\), the inequality

$$\begin{aligned} \mathop {\mathrm{ess\,sup\,}}\limits _{\xi \in \Xi ^{d}\left( x\right) }\mathbb {E}_{\mathbb {P}}\left[ \left. V\left( x+\left\langle \xi \left( \cdot \right) ,Y\left( \cdot \right) \right\rangle ,\cdot \right) \right| \fancyscript{G}\right] \le F\left( q,\cdot \right) \end{aligned}$$

holds a.s., thus we get

$$\begin{aligned} \mathop {\mathrm{ess\,sup\,}}\limits _{\xi \in \Xi ^{d}\left( x\right) }\mathbb {E}_{\mathbb {P}}\left[ \left. V\left( x+\left\langle \xi \left( \cdot \right) ,Y\left( \cdot \right) \right\rangle ,\cdot \right) \right| \fancyscript{G}\right] \le G\left( x,\cdot \right) \,\text {a.s.} \end{aligned}$$

It remains to verify that the reverse inequality is also true (except possibly on a set of measure zero). This will be achieved in three steps.

(a) :

Let us start by taking a strictly decreasing sequence \(\left\{ q_{n}\text {; }n\in \mathbb {N}\right\} \) of rational numbers satisfying \(x<q_{n}<x+1\) and \(\lim _{n\rightarrow +\infty } q_{n}=x\). Now, given any \(n\in \mathbb {N}\), it is straightforward that the family \( \left\{ \mathbb {E}_{\mathbb {P}}\left[ \left. V\left( q_{n}+\left\langle \xi \left( \cdot \right) ,Y\left( \cdot \right) \right\rangle ,\cdot \right) \right| \fancyscript{G}\right] \text {; }\xi \in \Xi ^{d}\left( q_{n}\right) \right\} \) is directed upwards, therefore one can find \(\zeta _n\in \Xi ^d(q_n)\) such that

$$\begin{aligned} \mathbb {E}_{\mathbb {P}}\left[ \left. V\left( q_{n}+\left\langle \zeta _{n}\left( \cdot \right) ,Y\left( \cdot \right) \right\rangle ,\cdot \right) \right| \fancyscript{G}\right] \ge F\left( q_{n},\cdot \right) -\frac{1}{n}\,\text {a.s.} \end{aligned}$$
(b) :

Next, fix an arbitrary \(n\). It was observed above that \(\zeta _{n}\in \Xi ^{d}\left( q_{n}\right) \subseteq \Xi ^{d}\left( x+1\right) \). Thus, taking \(\widehat{\zeta }_{n}\) to be its projection on \(D\), we know that

$$\begin{aligned} \mathbb {E}_{\mathbb {P}}\left[ \left. V\left( q_{n}+\left\langle \widehat{\zeta }_{n}\left( \cdot \right) ,Y\left( \cdot \right) \right\rangle ,\cdot \right) \right| \fancyscript{G}\right]&=\mathbb {E}_{\mathbb {P}}\left[ \left. V\left( q_{n}+\left\langle \zeta _{n}\left( \cdot \right) ,Y\left( \cdot \right) \right\rangle ,\cdot \right) \right| \fancyscript{G}\right] \nonumber \\&\ge F\left( q_{n},\cdot \right) -\frac{1}{n}\,\text {a.s.} \end{aligned}$$
(5.7)

Moreover, Lemma 4.7 allows us to conclude that \(||\widehat{\zeta }_{n}||\le K_{x+1}\) a.s. Therefore, we can extract a random subsequence \(\left\{ \widehat{\zeta }_{n_{k}};\,k\in \mathbb {N}\right\} \) such that \( \lim _{k\rightarrow +\infty } \widehat{\zeta }_{n_{k}}=\zeta \) a.s., for some \(\fancyscript{G}\)-measurable random variable \(\zeta \) (see Lemma 2 of [13]). But then

$$\begin{aligned} x+\left\langle \zeta \left( \omega \right) ,Y\left( \omega \right) \right\rangle =\lim _{k\rightarrow +\infty } \left( q_{n_{k}\left( \omega \right) }+\left\langle \widehat{\zeta }_{n_{k}\left( \omega \right) }\left( \omega \right) ,Y\left( \omega \right) \right\rangle \right) \ge 0 \end{aligned}$$

for \(\mathbb {P}\)-a.e. \(\omega \in \Omega \), i.e. \(\zeta \in \Xi ^{d}\left( x\right) \), which in turn implies that

$$\begin{aligned} \mathop {\mathrm{ess\,sup\,}}\limits _{\xi \in \Xi ^{d}\left( x\right) }\mathbb {E}_{\mathbb {P}}\left[ \left. V\left( x+\left\langle \xi \left( \cdot \right) ,Y\left( \cdot \right) \right\rangle ,\cdot \right) \right| \fancyscript{G}\right] \ge \mathbb {E}_{\mathbb {P}}\left[ \left. V\left( x+\left\langle \zeta \left( \cdot \right) ,Y\left( \cdot \right) \right\rangle ,\cdot \right) \right| \fancyscript{G}\right] \text {a.s.} \end{aligned}$$
(5.8)
(c) :

Finally, let us define the random variables \(f_{k}:~\Omega \rightarrow \mathbb {R}\) as follows,

$$\begin{aligned} f_{k}\left( \omega \right) \triangleq V\left( q_{n_{k}\left( \omega \right) }+\left\langle \widehat{\zeta }_{n_{k}\left( \omega \right) }\left( \omega \right) ,Y\left( \omega \right) \right\rangle ,\omega \right) ,\qquad \omega \in \Omega . \end{aligned}$$

By construction of the sequence \(\left\{ q_{n}{;} \ n\in \mathbb {N}\right\} \) and of the random subsequence \(\left\{ \widehat{\zeta }_{n_{k}}\text {; }k\in \mathbb {N}\right\} \), and by the continuity of the paths of \(V\) (see Assumption 4.3), it is clear that \( \lim _{k\rightarrow +\infty } f_{k}=V\left( x+\left\langle \zeta \left( \cdot \right) ,Y\left( \cdot \right) \right\rangle ,\cdot \right) \) a.s. We further observe that, for \(\mathbb {P}\)-a.e. \(\omega \in \Omega \),

$$\begin{aligned} f_{k}\left( \omega \right) \le V\left( x+1+\left\langle \widehat{\zeta }_{n_{k}\left( \omega \right) }\left( \omega \right) ,Y\left( \omega \right) \right\rangle ,\omega \right) \le \left[ \left( 1+x\right) ^{\overline{\gamma }}+1\right] L'\left( \omega \right) , \end{aligned}$$

where the first inequality follows from the monotonicity of \(V\) (again we refer to Assumption 4.3), and the second inequality is a simple consequence of Lemma 4.8 combined with the fact that

$$\begin{aligned} x+1+\left\langle \widehat{\zeta }_{n_{k}\left( \cdot \right) }\left( \cdot \right) ,Y\left( \cdot \right) \right\rangle \ge q_{n_{k}\left( \cdot \right) }+\left\langle \widehat{\zeta }_{n_{k}\left( \cdot \right) }\left( \cdot \right) ,Y\left( \cdot \right) \right\rangle \ge 0\,\text {a.s.} \end{aligned}$$

Hence, we may apply Fatou’s lemma (for the limit superior) to conclude that

$$\begin{aligned} \mathbb {E}_{\mathbb {P}}\left[ \left. V\left( x+\left\langle \zeta \left( \cdot \right) ,Y\left( \cdot \right) \right\rangle ,\cdot \right) \right| \fancyscript{G}\right]&\ge \limsup _{k\rightarrow +\infty } \mathbb {E}_{\mathbb {P}}\left[ \left. f_{k}\right| \fancyscript{G}\right] \nonumber \\&\ge \liminf _{k\rightarrow +\infty } \left[ F\left( q_{n_{k}\left( \cdot \right) },\cdot \right) -\frac{1}{n_k(\cdot )}\right] \ge \inf _{n\in \mathbb {N}} F\left( q_{n},\cdot \right) \,\text {a.s.,} \end{aligned}$$
(5.9)

from (5.7) applied on \(\{n_k=i\}\) for \(i \ge k\). Combining equations (5.8) and (5.9) finally gives the intended inequality

$$\begin{aligned} \mathop {\mathrm{ess\,sup\,}}\limits _{\xi \in \Xi ^{d}\left( x\right) }\mathbb {E}_{\mathbb {P}}\left[ \left. V\left( x+\left\langle \xi \left( \cdot \right) ,Y\left( \cdot \right) \right\rangle ,\cdot \right) \right| \fancyscript{G}\right] \ge \inf _{n\in \mathbb {N}} F\left( q_{n},\cdot \right) \ge G\left( x,\cdot \right) \,\text {a.s.} \end{aligned}$$
(iii) :

Thirdly, \(G\) is, by construction, right-continuous. Moreover it is easy to see that

$$\begin{aligned} G(x,\omega )=1_A(\omega ) \inf _{n \in \mathbb {N}} \{F(r_n,\omega )1_{r_n >x} +(+\infty ) 1_{r_n \le x}\} \end{aligned}$$

where \(\{r_n; \, n \in \mathbb {N}\}\) is an enumeration of \(\mathbb {Q}\) (we make the usual convention that \(0 \times \infty = 0\)). Recalling that \(\omega \rightarrow F(r_n,\omega )\) is \(\fancyscript{G}\)-measurable for each \(r_n\), it follows that \(G\) is measurable with respect to the product \(\sigma \)-algebra \(\fancyscript{B}\left( \left[ \left. 0,+\infty \right) \right. \right) \otimes \fancyscript{G}\).

(iv) :

Now consider an arbitrary \(\fancyscript{G}\)-measurable random variable \(H\ge 0\) a.s. We wish to show that

$$\begin{aligned} G\left( H\left( \cdot \right) ,\cdot \right) =\mathop {\mathrm{ess\,sup\,}}\limits _{\xi \in \Xi ^{d}\left( H\right) }\mathbb {E}_{\mathbb {P}}\left[ \left. V\left( H\left( \cdot \right) +\left\langle \xi \left( \cdot \right) ,Y\left( \cdot \right) \right\rangle ,\cdot \right) \right| \fancyscript{G}\right] \text{ a.s. } \end{aligned}$$
(5.10)

We have just proved above that (5.10) holds true for a non-negative constant \(H\). It is easy to show that it holds also true for \(\fancyscript{G}\)-measurable countable step-functions \(H\). Next, suppose \(H\) is any bounded, \(\fancyscript{G}\)-measurable, non-negative (a.s.) random variable, so there exists some constant \(M>0\) such that \(H\le M\) a.s. It is a well-known fact that we can take a non-increasing sequence \(\left\{ H_{n}\text {;}\,n\in \mathbb {N}\right\} \) of \(\fancyscript{G}\)-measurable step-functions converging to \(H\) a.s., and such that, for every \(n\in \mathbb {N}\), \(H_{n}\le M\) a.s. Then, fixing an arbitrary \(\xi \in \Xi ^{d}\left( H\right) \), we have for every \(n\in \mathbb {N}\) that \( H_{n}+\left\langle \xi ,Y\right\rangle \ge H+\left\langle \xi ,Y\right\rangle \ge 0\) a.s., therefore

$$\begin{aligned} G\left( H_{n}\left( \cdot \right) ,\cdot \right)&=\mathop {\mathrm{ess\,sup\,}}\limits _{\zeta \in \Xi ^{d}\left( H_{n}\right) }\mathbb {E}_{\mathbb {P}}\left[ \left. V\left( H_{n}\left( \cdot \right) +\left\langle \zeta \left( \cdot \right) ,Y\left( \cdot \right) \right\rangle ,\cdot \right) \right| \fancyscript{G}\right] \\&\ge \mathbb {E}_{\mathbb {P}}\left[ \left. V\left( H_{n}\left( \cdot \right) +\left\langle \xi \left( \cdot \right) ,Y\left( \cdot \right) \right\rangle ,\cdot \right) \right| \fancyscript{G}\right] \,\text {a.s.} \end{aligned}$$

(recall that (5.10) is true for step-functions), which in turn yields

$$\begin{aligned} \liminf _{n\rightarrow +\infty }G\left( H_{n}\left( \cdot \right) ,\cdot \right) \ge \liminf _{n\rightarrow +\infty } \mathbb {E}_{\mathbb {P}}\left[ \left. V\left( H_{n}\left( \cdot \right) +\left\langle \xi \left( \cdot \right) ,Y\left( \cdot \right) \right\rangle ,\cdot \right) \right| \fancyscript{G}\right] \,\text {a.s.} \end{aligned}$$

On the one hand, we get by right-continuity of \(G\) that \(\lim _{n\rightarrow +\infty }G\left( H_{n}\left( \cdot \right) ,\cdot \right) =G\left( H\left( \cdot \right) ,\cdot \right) \) a.s. On the other hand, we can apply Fatou’s lemma (for the limit inferior, see Assumption 4.5) to conclude

$$\begin{aligned} \liminf _{n\rightarrow +\infty } \mathbb {E}_{\mathbb {P}}\left[ \left. V\left( H_{n}\left( \cdot \right) +\left\langle \xi \left( \cdot \right) ,Y\left( \cdot \right) \right\rangle ,\cdot \right) \right| \fancyscript{G}\right] \ge \mathbb {E}_{\mathbb {P}}\left[ \left. V\left( H\left( \cdot \right) +\left\langle \xi \left( \cdot \right) ,Y\left( \cdot \right) \right\rangle ,\cdot \right) \right| \fancyscript{G}\right] \,\text {a.s.}, \end{aligned}$$

hence \(\mathop {\mathrm{ess\,sup\,}}\limits _{\xi \in \Xi ^{d}\left( H\right) }\mathbb {E}_{\mathbb {P}}\left[ \left. V\left( H\left( \cdot \right) +\left\langle \xi \left( \cdot \right) ,Y\left( \cdot \right) \right\rangle ,\cdot \right) \right| \fancyscript{G}\right] \le G\left( H\left( \cdot \right) ,\cdot \right) \) a.s. (by the arbitrariness of \(\xi \in \Xi ^{d}\left( H\right) \)). Now, to prove the reverse inequality, we can construct (as in part (ii) of this proof) a sequence \(\left\{ \zeta _{n}\text {;}\,n\in \mathbb {N}\right\} \) such that, for every \(n\in \mathbb {N}\), we have \(\zeta _{n}\in \Xi ^{d}\left( H_{n}\right) \), \(\zeta _{n}\left( \omega \right) \in D\left( \omega \right) \) for \(\mathbb {P}\)-a.e. \(\omega \in \Omega \), and

$$\begin{aligned} G\left( H_{n}\left( \cdot \right) ,\cdot \right) -\frac{1}{n}&=\mathop {\mathrm{ess\,sup\,}}\limits _{\xi \in \Xi ^{d}\left( H_{n}\right) }\mathbb {E}_{\mathbb {P}}\left[ \left. V\left( H_{n}\left( \cdot \right) +\left\langle \xi \left( \cdot \right) ,Y\left( \cdot \right) \right\rangle ,\cdot \right) \right| \fancyscript{G}\right] -\frac{1}{n}\nonumber \\&\le \mathbb {E}_{\mathbb {P}}\left[ \left. V\left( H_{n}\left( \cdot \right) +\left\langle \zeta _{n}\left( \cdot \right) ,Y\left( \cdot \right) \right\rangle ,\cdot \right) \right| \fancyscript{G}\right] \,\text {a.s.} \end{aligned}$$
(5.11)

We remark further that each \(\zeta _{n}\) belongs to \(\Xi ^{d}\left( M\right) \) (because \(M+\left\langle \zeta _{n},Y\right\rangle \ge H_{n}+\left\langle \zeta _{n},Y\right\rangle \ge 0\) a.s.), so by Lemma 4.7 there exists a random variable \(K_{M}\) such that \(\left\| \zeta _{n}\right\| \le K_{M}\) a.s. Therefore we can select a random subsequence \(\left\{ \zeta _{n_{k}}\text {; }k\in \mathbb {N}\right\} \) with \( \lim _{k\rightarrow +\infty }\zeta _{n_{k}}=\zeta \) a.s., for some \(\fancyscript{G}\)-measurable \(\zeta \). Clearly,

$$\begin{aligned} H+\left\langle \zeta ,Y\right\rangle =\lim _{k\rightarrow +\infty }\left( H_{n_{k}}+\left\langle \zeta _{n_{k}},Y\right\rangle \right) \,\text {a.s.}, \end{aligned}$$

and for every \(k\in \mathbb {N}\),

$$\begin{aligned} H_{n_{k}}+\left\langle \zeta _{n_{k}},Y\right\rangle =\sum _{i=k}^{+\infty }\left( H_{i}+\left\langle \zeta _{i},Y\right\rangle \right) 11_{\left\{ n_{k}\left( \cdot \right) =i\right\} }\ge 0\,\text {a.s.}, \end{aligned}$$

hence \(\zeta \in \Xi ^{d}\left( H\right) \). Consequently,

$$\begin{aligned} \mathop {\mathrm{ess\,sup\,}}\limits _{\xi \in \Xi ^{d}\left( H\right) }\mathbb {E}_{\mathbb {P}}\left[ \left. V\left( H\left( \cdot \right) +\left\langle \xi \left( \cdot \right) ,Y\left( \cdot \right) \right\rangle ,\cdot \right) \right| \fancyscript{G}\right] \ge \mathbb {E}_{\mathbb {P}}\left[ \left. V\left( H\left( \cdot \right) +\left\langle \zeta \left( \cdot \right) ,Y\left( \cdot \right) \right\rangle ,\cdot \right) \right| \fancyscript{G}\right] \,\text {a.s.,} \end{aligned}$$

by definition of essential supremum. Besides, we have by Lemma 4.8 that, for every \(k\in \mathbb {N}\),

$$\begin{aligned} V^{+}\left( H_{n_{k}\left( \cdot \right) }\left( \cdot \right) +\left\langle \zeta _{n_{k}\left( \cdot \right) }\left( \cdot \right) ,Y\left( \cdot \right) \right\rangle ,\cdot \right) \le V^{+}\left( M+\left\langle \zeta _{n_{k}\left( \cdot \right) }\left( \cdot \right) ,Y\left( \cdot \right) \right\rangle ,\cdot \right) \le L_{M}\left( \cdot \right) \,\text {a.s.} \end{aligned}$$

(note that \(\zeta _{n_{k}}\in \Xi ^{d}\left( M\right) \)), so the limsup Fatou lemma yields (cf. Assumption 4.4)

$$\begin{aligned}&\mathbb {E}_{\mathbb {P}}\left[ \left. V\left( H\left( \cdot \right) +\left\langle \zeta \left( \cdot \right) ,Y\left( \cdot \right) \right\rangle ,\cdot \right) \right| \fancyscript{G}\right] \\&\quad \ge \limsup _{k\rightarrow +\infty }\mathbb {E}_{\mathbb {P}}\left[ \left. V\left( H_{n_{k}\left( \cdot \right) }\left( \cdot \right) +\left\langle \zeta _{n_{k}\left( \cdot \right) }\left( \cdot \right) ,Y\left( \cdot \right) \right\rangle ,\cdot \right) \right| \fancyscript{G}\right] \\&\quad \ge \limsup _{k\rightarrow +\infty }\left( G\left( H_{n_{k}\left( \cdot \right) }\left( \cdot \right) ,\cdot \right) -\frac{1}{n_{k}\left( \cdot \right) } \right) =G\left( H\left( \cdot \right) ,\cdot \right) \,\text {a.s.,} \end{aligned}$$

where the last inequality follows from (5.11). Combining the inequalities above, we establish (4.7) for any bounded \(H\) as well. Finally, we extend the above result to an arbitrary \(\fancyscript{G}\)-measurable \(H\ge 0\) a.s. Since \(H=\sum _{n\in \mathbb {N}}H_{n}\), with each \(H_{n}\triangleq H11_{\left\{ n-1\le H<n\right\} }\) \(\fancyscript{G}\)-measurable and bounded, we can obtain the desired equality from the bounded case.

(v) :

Lastly, we claim that almost all paths of \(G\) are left-continuous. This will be done by constructing some left-continuous function \(\overline{G}\) such that

$$\begin{aligned} \mathbb {P}\left\{ \omega \in {\Omega }\text {: }\forall \,x\ge 0,\ G\left( x,\omega \right) =\overline{G}\left( x,\omega \right) \right\} =1. \end{aligned}$$

We define

$$\begin{aligned} \overline{G}\left( x,\omega \right) \triangleq \left\{ \begin{array}{ll} \sup _{\begin{array}{c} q\in \mathbb {Q}\\ q<x \end{array}} G\left( q,\omega \right) , &{}\quad \text {if }x>0,\\ G\left( 0,\omega \right) , &{}\quad \text {otherwise.} \end{array} \right. \end{aligned}$$

From (iii) recall that \(G\) is \(\fancyscript{B}\left( \left[ \left. 0,+\infty \right) \right. \right) \otimes \fancyscript{G}\)-measurable, so it is obvious that \(\overline{G}\) is \(\fancyscript{B}\left( \left[ \left. 0,+\infty \right) \right. \right) \otimes \fancyscript{G}\)-measurable too. Besides, it is trivial to check that, for every \(\omega \in \Omega \), the function \(\overline{G}\left( \cdot ,\omega \right) \) is non-decreasing on \(\left( 0,+\infty \right) \). We remark further that, by construction, all paths of \(\overline{G}\) are left-continuous on \(\left( 0,+\infty \right) \). It follows immediately from the monotonicity of all the sample paths of \(G\) that the inequality \(G\left( x,\omega \right) \ge \overline{G}\left( x,\omega \right) \) holds true simultaneously for every \(x\ge 0\), for every \(\omega \in \Omega \). In particular, this gives that, for every \(\omega \in \Omega \) and for all \(x\ge 0\), it holds that \(\overline{G}\left( x,\omega \right) <+\infty \). At last, we shall show that \(\mathbb {P}\left\{ \omega \in {\Omega }\text {: }\forall \,x\ge 0,\ G\left( x,\omega \right) =\overline{G}\left( x,\omega \right) \right\} =1\).

(a) :

The proof is by contradiction. Let us suppose that the set

$$\begin{aligned} \Omega _{1}\triangleq \left\{ \omega \in {\Omega }\,\text {:}\,\exists \,x>0\,\text {s.t.}\,G\left( x,\omega \right) >\overline{G}\left( x,\omega \right) \right\} \end{aligned}$$

has strictly positive measure, i.e., \(\mathbb {P}\left( \Omega _{1}\right) >0\). Note that, because \(\left( \Omega ,\fancyscript{G},\mathbb {P}\right) \) is a complete measure space, we can apply the measurable projection theorem (see e.g. Theorem 4 in Sainte-Beuve [22]) to deduce that Footnote 3 \(\Omega _{1}={\text {Proj}}_{\Omega }\!\left( \left( G-\overline{G}\right) ^{-1}\left( \left( 0,+\infty \right) \right) \right) \) belongs to \(\fancyscript{G}\). Consider the multi-function \(\fancyscript{E}:~\Omega \rightrightarrows \left[ \left. 0,+\infty \right) \right. \) given by

$$\begin{aligned} \fancyscript{E}\left( {\omega }\right) \triangleq \left\{ \begin{array}{ll} \left\{ x>0\text {: }G\left( x,\omega \right) >\overline{G}\left( x,\omega \right) \right\} , &{}\quad \text {if } \omega \in \Omega _{1},\\ \{1\}, &{}\quad \text {otherwise.} \end{array} \right. \end{aligned}$$

Its graph is given by

$$\begin{aligned} {{\mathrm{gph}}}\fancyscript{E}=\left( \Omega _{1}^{c}\times \left\{ 1\right\} \right) \cup \left( \left[ \Omega _{1}\times \left( 0,+\infty \right) \right] \cap \left[ \left( G-\overline{G}\right) ^{-1}\left( \left( 0,+\infty \right) \right) \right] \right) \end{aligned}$$

and belongs to \(\fancyscript{B}\left( \left[ \left. 0,+\infty \right) \right. \right) \otimes \fancyscript{G}\). Consequently, we can apply the von Neumann-Aumann theorem (see e.g. Theorem 3 in Sainte-Beuve [22]) to produce a \(\fancyscript{G}\)-measurable selector \(H{:}~{\Omega }\rightarrow \left[ \left. 0,+\infty \right) \right. \) of \(\fancyscript{E}\). In particular, this implies that

$$\begin{aligned} \mathbb {P}\left\{ \omega \in {\Omega }\text {: }G\left( H\left( \omega \right) ,\omega \right) >\overline{G}\left( H\left( \omega \right) ,\omega \right) \right\} \ge \mathbb {P}\left( \Omega _{1}\right) >0. \end{aligned}$$
(5.12)

As \(G\left( 0,\omega \right) =\overline{G}\left( 0,\omega \right) \), we get that \(H>0\). Furthermore, we may and shall assume, without loss of generality, that there exists some \(\varepsilon \in \left. \left( 0,1\right. \right] \) such that \(H>\varepsilon \).

(b):

On the other hand, we shall see that \(G\left( H\left( \omega \right) ,\omega \right) \le \overline{G}\left( H\left( \omega \right) ,\omega \right) \) holds for \(\mathbb {P}\)-a.e. \(\omega \in \Omega \), contradicting (5.12). Firstly, fix an arbitrary \(n\in \mathbb {N}\). As in part (ii) of this proof, it is possible to construct some \(\zeta _{n}\in \Xi ^{d}\left( H\right) \) such that, for \(\mathbb {P}\)-a.e. \(\omega \in \Omega \),

$$\begin{aligned} \mathbb {E}_{\mathbb {P}}\left[ \left. V\left( H\left( \cdot \right) +\left\langle \zeta _{n}\left( \cdot \right) ,Y\left( \cdot \right) \right\rangle ,\cdot \right) \right| \fancyscript{G}\right] \left( \omega \right) \ge G\left( H\left( \omega \right) ,\omega \right) -\frac{1}{n}. \end{aligned}$$
(5.13)

Next, setting for every \(m\in \mathbb {N}\) (recall that \(H>\varepsilon \)),

$$\begin{aligned} f_{n}^{m}\left( \omega \right) \triangleq V\left( H\left( \omega \right) -\frac{\varepsilon }{m}+\frac{H\left( \omega \right) -\varepsilon /m}{H\left( \omega \right) }\left\langle \zeta _{n}\left( \omega \right) ,Y\left( \omega \right) \right\rangle ,\omega \right) ,\quad \omega \in \Omega , \end{aligned}$$

it is trivial by continuity (see Assumption 4.3) that \(\left\{ f_{n}^{m}\text {;}\,m\in \mathbb {N}\right\} \) converges a.s. to \(V\left( H\left( \cdot \right) +\left\langle \zeta _{n}\left( \cdot \right) ,Y\left( \cdot \right) \right\rangle ,\cdot \right) \) as \(m\rightarrow +\infty \). Thus, Fatou’s lemma gives

$$\begin{aligned} \liminf _{m\rightarrow +\infty }\mathbb {E}_{\mathbb {P}}\left[ \left. \left[ f_{n}^{m}\right] ^{+}\right| \fancyscript{G}\right] \ge \mathbb {E}_{\mathbb {P}}\left[ \left. V^{+}\left( H\left( \cdot \right) +\left\langle \zeta _{n}\left( \cdot \right) ,Y\left( \cdot \right) \right\rangle ,\cdot \right) \right| \fancyscript{G}\right] \,\text {a.s.} \end{aligned}$$

Secondly, we note that, for each \(m\in \mathbb {N}\), the random vector \(\zeta _{n}\left( H-\varepsilon /m\right) /H\) belongs to \(\Xi ^{d}\left( H-\varepsilon /m\right) \), because

$$\begin{aligned} H-\frac{\varepsilon }{m}+\left\langle \frac{H-\varepsilon /m}{H}\zeta _{n},Y \right\rangle =\frac{H-\varepsilon /m}{H}\left( H+\left\langle \zeta _{n},Y \right\rangle \right) \ge 0\,\text {a.s.} \end{aligned}$$

(recall that \(H>\varepsilon \) and \(\zeta _{n}\in \Xi ^{d}\left( H\right) \)). Therefore, given Assumption 4.5 and the fact that, for every \(m\in \mathbb {N}\), the inequality \(\left[ f_{n}^{m}\right] ^{-}\le V^{-}\left( 0,\cdot \right) \) is true a.s., we can apply the limsup Fatou lemma to obtain

$$\begin{aligned} \limsup _{m\rightarrow +\infty }\mathbb {E}_{\mathbb {P}}\left[ \left. \left[ f_{n}^{m}\right] ^{-}\right| \fancyscript{G}\right] \le \mathbb {E}_{\mathbb {P}}\left[ \left. V^{-}\left( H\left( \cdot \right) +\left\langle \zeta _{n}\left( \cdot \right) ,Y\left( \cdot \right) \right\rangle ,\cdot \right) \right| \fancyscript{G}\right] \,\text {a.s.} \end{aligned}$$

Combining both inequalities yields

$$\begin{aligned} \liminf _{m\rightarrow +\infty }\mathbb {E}_{\mathbb {P}}\left[ \left. f_{n}^{m}\right| \fancyscript{G}\right] \ge \mathbb {E}_{\mathbb {P}}\left[ \left. V\left( H\left( \cdot \right) +\left\langle \zeta _{n}\left( \cdot \right) ,Y\left( \cdot \right) \right\rangle ,\cdot \right) \right| \fancyscript{G}\right] \,\text {a.s.,} \end{aligned}$$

and from (5.13) we get

$$\begin{aligned} \liminf _{m\rightarrow +\infty }\mathbb {E}_{\mathbb {P}}\left[ \left. f_{n}^{m}\right| \fancyscript{G}\right] \ge G\left( H\left( \cdot \right) ,\cdot \right) -\frac{1}{n}\,\text {a.s.} \end{aligned}$$
(5.14)

Besides, for every \(m\in \mathbb {N}\) we have that

$$\begin{aligned} \mathop {\mathrm{ess\,sup\,}}\limits _{\xi \in \Xi ^{d}\left( H-\varepsilon /m\right) }\mathbb {E}_{\mathbb {P}}\left[ \left. V\left( H\left( \cdot \right) -\frac{\varepsilon }{m}+\left\langle \xi \left( \cdot \right) ,Y\left( \cdot \right) \right\rangle ,\cdot \right) \right| \fancyscript{G}\right] \ge \mathbb {E}_{\mathbb {P}}\left[ \left. f_{n}^{m}\right| \fancyscript{G}\right] \,\text {a.s.,} \end{aligned}$$

and so

$$\begin{aligned}&\liminf _{m\rightarrow +\infty }\mathop {\mathrm{ess\,sup\,}}\limits _{\xi \in \Xi ^{d}\left( H-\varepsilon /m\right) }\mathbb {E}_{\mathbb {P}}\left[ \left. V\left( H\left( \cdot \right) -\frac{\varepsilon }{m}+\left\langle \xi \left( \cdot \right) ,Y\left( \cdot \right) \right\rangle ,\cdot \right) \right| \fancyscript{G}\right] \\&\quad \ge \liminf _{m\rightarrow +\infty }\mathbb {E}_{\mathbb {P}}\left[ \left. f_{n}^{m}\right| \fancyscript{G}\right] \text { a.s.} \end{aligned}$$

On the other hand, let \(m\in \mathbb {N}\) be arbitrary, but fixed. Then we know by the preceding step that

$$\begin{aligned} \mathop {\mathrm{ess\,sup\,}}\limits _{\xi \in \Xi ^{d}\left( H-\varepsilon /m\right) }\mathbb {E}_{\mathbb {P}}\left[ \left. V\left( H\left( \cdot \right) -\frac{\varepsilon }{m}+\left\langle \xi \left( \cdot \right) ,Y\left( \cdot \right) \right\rangle ,\cdot \right) \right| \fancyscript{G}\right] \left( \omega \right) =G\left( H\left( \omega \right) -\frac{\varepsilon }{m},\omega \right) \end{aligned}$$

for every \(\omega \) outside a \(\mathbb {P}\)-null set. Using (5.14), we get

$$\begin{aligned} \liminf _{m\rightarrow +\infty }G\left( H\left( \cdot \right) -\frac{\varepsilon }{m},\cdot \right) \ge G\left( H\left( \cdot \right) ,\cdot \right) -\frac{1}{n}\text { a.s.} \end{aligned}$$
(5.15)

Next, choosing \(q_{m}(\omega )\in \mathbb {Q}\), \(q_m(\omega )>0\) such that \(H\left( \omega \right) -\varepsilon /m\le q_{m}(\omega )<H\left( \omega \right) \), it follows immediately from the definition of \(\overline{G}\) (recall that \(H>\varepsilon >0\)) and from the monotonicity of \(G\) (see the first part of this proof) that

$$\begin{aligned} \overline{G}\left( H\left( \omega \right) ,\omega \right)&=\sup _{\begin{array}{c} q\in \mathbb {Q}\\ q<H\left( \omega \right) \end{array}} G\left( q,\omega \right) \ge G\left( q_{m}(\omega ),\omega \right) \\&\ge G\left( H\left( \omega \right) -\varepsilon /m,\omega \right) \ge \inf _{k\ge m}G\left( H\left( \omega \right) -\varepsilon /k,\omega \right) , \end{aligned}$$

consequently,

$$\begin{aligned} \overline{G}\left( H\left( \cdot \right) ,\cdot \right) \ge \sup _{m\in \mathbb {N}}\inf _{k\ge m}G\left( H\left( \cdot \right) -\varepsilon /k,\cdot \right) =\liminf _{m\rightarrow +\infty }G\left( H\left( \cdot \right) -\varepsilon /m,\cdot \right) \text { a.s.} \end{aligned}$$

So, from (5.15), for every \(n\in \mathbb {N}\), \( \overline{G}\left( H\left( \cdot \right) ,\cdot \right) \ge G\left( H\left( \cdot \right) ,\cdot \right) -1/n\) a.s., hence

$$\begin{aligned} \overline{G}\left( H\left( \cdot \right) ,\cdot \right) \ge \limsup _{n\rightarrow +\infty }\left( G\left( H\left( \cdot \right) ,\cdot \right) -\frac{1}{n}\right) =G\left( H\left( \cdot \right) ,\cdot \right) \text { a.s.}, \end{aligned}$$

as claimed.\(\square \)

Proof of Proposition 4.10

As in (ii) of the previous proof, one can show that there exists some sequence \(\xi _{n}\left( \cdot \right) \in \Xi ^{d}\left( H\right) \) that attains the essential supremum in (4.7). We may assume that \(\xi _{n}\left( \cdot \right) \in D(\cdot )\) a.s., and hence for every \(n\in \mathbb {N}\) we have by Lemma 4.7 that \(\left\| \xi _{n}\left( \cdot \right) \right\| \le H\left( \cdot \right) /\beta \left( \cdot \right) \) a.s. Next, Lemma 2 of Kabanov and Stricker [13] implies the existence of a \(\fancyscript{G}\)-measurable random subsequence \(\left\{ \xi _{n_{k}}\text {;}\,k\in \mathbb {N}\right\} \) such that \(\lim _{k\rightarrow +\infty }\xi _{n_{k}\left( \cdot \right) }\left( \cdot \right) =\xi \left( \cdot \right) \) a.s. for some \(\xi \left( \cdot \right) \in \Xi ^{d}\left( H\right) \). Lemmata 4.8 and 4.9 allow the use of the (conditional) Fatou lemma, hence we get that \(\widetilde{\xi }\left( H\right) \left( \cdot \right) \triangleq \xi \left( \cdot \right) \) is as claimed.\(\square \)

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Carassus, L., Rásonyi, M. & Rodrigues, A.M. Non-concave utility maximisation on the positive real axis in discrete time. Math Finan Econ 9, 325–349 (2015). https://doi.org/10.1007/s11579-015-0146-4

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