By definition of the \(\operatorname{ae}\) filter and of a supremum of filters.

Let \(Q \in \mathcal{Q}\). Since \(f\) is an E-variable for \(\mathcal{Q}\), we have \(Q[f] \le 1\). Hence \(Q[\{ f = \infty \} ] = 0\).

Since \(\mathcal{E}_{\mathcal{Q}} \subseteq \mathcal{E}^R_{\mathcal{Q}}\), the left hand side is greater than or equal to the right hand side.

For the other direction, let \(\eta \in \mathcal{E}^R_{\mathcal{Q}}\) be a randomized E-variable. By concavity of \(U\),

The function \(x \mapsto \eta (x)[\operatorname{id}]\) is a measurable function and belongs to \(\mathcal{E}_{\mathcal{Q}}\).

The function \(f : \mathcal{X} \to \mathbb {R}_{+,\infty }\) defined by \(f(x) = +\infty \) if \(x \in A\) and \(f(x) = 0\) otherwise is an E-variable for \(\mathcal{Q}\) since \(Q[f] = 0\) for all \(Q \in \mathcal{Q}\).

Let \(R_1, R_2 \in \mathcal{P}(\mathcal{X})\) and let \(\theta \in [0,1]\).

For \(Q \in \mathcal{Q}\),

For \(Q \in \mathcal{Q}\),

By Lemma 2.22, \(\{ \xi \circ \kappa \mid \xi \in \mathcal{E}^R_{\kappa \circ \mathcal{Q}}\} \subseteq \mathcal{E}^R_{\mathcal{Q}}\), hence

We can use Lemma 2.18 to obtain the data-processing with a Markov kernel for e-variables from the result for randomized e-variables (Theorem 2.23).

Use Lemma 2.24 with the deterministic Markov kernel given by \(f\).

To prove the first inequality, we use Lemma 2.25:

For the reverse inequality,

Use Lemma 2.31 with the e-variable \(f = 1\).

For \(f \in \mathcal{E}_{\mathcal{Q}_2}\), we have \(f \in \mathcal{E}_{\mathcal{Q}_1}\) since \(\mathcal{Q}_1 \subseteq \mathcal{Q}_2\). Then \(P\left[f/f^*\right] \le 1\) by the numeraire property for \((P, \mathcal{Q}_1)\). Thus \(f^*\) is a numeraire for \((P, \mathcal{Q}_2)\).

Proved in the Brownian motion project.

Let \((f_n)_{n \in \mathbb {N}}\) be a sequence of e-variables in \(\mathcal{E}_{\mathcal{Q}}\) such that \(n \mapsto P[U(f_n)]\) is non-decreasing and \(P[U(f_n)] \to \sup _{f \in \mathcal{E}_{\mathcal{Q}}} P[U(f)]\). Let \(\tilde{f}_n\) be a sequence of measurable functions and let \(\tilde{f} : \mathcal{X} \to \mathbb {R}_{+, \infty }\), with \(\tilde{f}_n \in \mathrm{conv}(f_n, f_{n+1}, \ldots )\) such that \(\tilde{f}_n \to \tilde{f}\) \(P\)-almost surely. That sequence exists by Lemma 2.35. By convexity of \(\mathcal{E}_{\mathcal{Q}}\), \(\tilde{f}_n \in \mathcal{E}_{\mathcal{Q}}\) for all \(n\).

Let \(f^* = \liminf _{n \to \infty } \tilde{f}_n\). Note that \(f^*\) is a measurable function such that \(f^* = \tilde{f}\) \(P\)-almost surely. We have \(f^* \in \mathcal{E}_{\mathcal{Q}}\) by an application of Fatou’s lemma: for \(Q \in \mathcal{Q}\),

By concavity of \(U\) and the fact that \(P[U(f_n)]\) is non-decreasing, we have

We also have \(f^* = \limsup _{n \to \infty } \tilde{f}_n\) \(P\)-almost surely, hence by Fatou’s lemma (using that \(U\) is bounded from above),

Since \(f^* \in \mathcal{E}_{\mathcal{Q}}\), we also have \(P[U(f^*)] \le \sup _{f \in \mathcal{E}_{\mathcal{Q}}} P[U(f)]\), which proves the equality.

Let \(f_0\) be as in Lemma 2.36. The set \(\{ f_0 = \infty \} \) is \(\mathcal{Q}\)-null by Lemma 2.12.

TODO

For \(f \in \mathcal{E}_{\mathcal{Q}}\), let \(f_t = t f + (1-t) f^*\) for \(t \in [0,1]\). By concavity of \(U\), \(U(f_t) \ge t U(f) + (1-t) U(f^*)\), hence

TODO

We first need to check that \(E^*_{P_1, \mathcal{Q}_1} E^*_{P_2, \mathcal{Q}_2}\) is in \(\mathcal{E}_{\mathcal{Q}_1 \times \mathcal{Q}_2}\). For \(Q_1 \in \mathcal{Q}_1\) and \(Q_2 \in \mathcal{Q}_2\), we have

Let’s now check the numeraire property. Let \(f \in \mathcal{E}_{\mathcal{Q}_1 \times \mathcal{Q}_2}\). Let \(E^*_1 = E^*_{P_1, \mathcal{Q}_1}\) and \(E^*_2 = E^*_{P_2, \mathcal{Q}_2}\).

If we can show that \(x \mapsto P_2\left[\frac{f(x, \cdot )}{E^*_2}\right]\) is in \(\mathcal{E}_{\mathcal{Q}_1}\), then we can apply the definition of the numeraire for \((P_1, \mathcal{Q}_1)\) to obtain

Let then \(Q_1 \in \mathcal{Q}_1\). We want to show that \(Q_1 \left[P_2\left[\frac{f(x, \cdot )}{E^*_2}\right]\right] \le 1\). By Fubini’s theorem, we have

It now suffices to show that \(y \mapsto Q_1 \left[f(\cdot , y)\right]\) is in \(\mathcal{E}_{\mathcal{Q}_2}\). Let \(Q_2 \in \mathcal{Q}_2\). Then

The last inequality holds since \(f \in \mathcal{E}_{\mathcal{Q}_1 \times \mathcal{Q}_2}\).

For \(Q \in \mathcal{Q}\) and \(f \in \mathcal{E}_{\mathcal{Q}\otimes H}\), \(Q[x \mapsto \eta (x)[f]] = (Q \otimes \eta )[f] \le 1\), hence \((x \mapsto \eta (x)[f(x, \cdot )]) \in \mathcal{E}_{\mathcal{Q}}\).

First we need to show that \(E^*_{P, \mathcal{Q}}\) is in \(\mathcal{E}_{\mathcal{Q} \otimes \kappa }\). It is measurable and for \(Q \in \mathcal{Q}\),

We then check the numeraire condition. By Lemma 2.41, \((x \mapsto \kappa (x)[f]) \in \mathcal{E}_{\mathcal{Q}}\). By the definition of the numeraire, we then have

That is, \((P \otimes \kappa )\left[\frac{f}{E^*_{P, \mathcal{Q}}}\right] \le 1\) for all \(f \in \mathcal{E}_{\mathcal{Q}\otimes \kappa }\). We have proved that \(E^*_{P, \mathcal{Q}}\) is the numeraire for \((P \otimes \kappa , \mathcal{Q} \otimes \kappa )\).

Let \(f \in \mathcal{E}_{\mathcal{Q} \otimes H}\).

It suffices to show that \(f/E^*_{P, \mathcal{Q}}\) is in \(\mathcal{E}_{P \otimes H}\) to obtain that this is less than 1 by the numeraire property. Let \(\eta \in H\). By Lemma 2.42,

and since \(f \in \mathcal{E}_{\mathcal{Q} \otimes H}\) it is in \(\mathcal{E}_{\mathcal{Q} \otimes \eta }\) and the numeraire property gives that this ratio is less than or equal to 1. We conclude that \(E^*_{P, \mathcal{Q}}E^*_{P \otimes \kappa , P \otimes H}\) satisfies the numeraire property.

We then have

We conclude that

Taking the supremum over \(f \in \mathcal{E}_{\mathcal{Q} \otimes H}\), we obtain the desired inequality.

\(1/E^*_{\kappa , \{ \eta \} }\) is jointly measurable and satisfies for all \(x \in \mathcal{X}\), \(1/E^*_{\kappa , \{ \eta \} }(x, \cdot ) =_{a.e.} \frac{d\eta (x)}{d \kappa (x)}\) since that’s the numeraire for a singleton. That’s the definition of a Radon-Nikodym derivative of kernels.

\(E^*_{\kappa , H}\) is a numeraire for \((P \otimes \kappa , \{ P\} \otimes H)\), so we get the numeraire property as in Theorem 2.43. We now prove that \(E^*_{P, \mathcal{Q}}E^*_{\kappa , H} \in \mathcal{E}_{\mathcal{Q} \otimes H}\). Let \(Q \in \mathcal{Q}\) and \(\eta \in H\).

For all \(x\), \(\eta (x)\left[E^*_{\kappa (x), H(x)}\right] \le 1\), since \(E^*_{\kappa (x), H(x)}\) is an e-variable for \(H(x)\).

For each \(x \in \mathcal{X}\), we can apply Theorem 2.39 to the Markov kernel \(\kappa (x)\) and the set of kernels \(H(x)\) to obtain a numeraire \(E^*_{\kappa (x), H(x)}\). If \(\mathcal{X}\) is countable, we can simply take the function \(E^*_{\kappa , H}(x,y) = E^*_{\kappa (x), H(x)}(y)\), and the countability ensures the joint measurability.

The case of countable \(\mathcal{X}\) is covered by Lemma 2.48. (TODO: only for measurable singletons though)

TODO: \(\kappa \ll H\) hyp?

Let’s now consider the case where \(\mathcal{Y}\) is countably generated. Then there exists a sequence of finer and finer partitions of \(\mathcal{Y}\) into finitely many measurable sets \((A_{n,m})_{n \in \mathbb {N}, m \le 2^n}\) with \(\mathcal{Y} = \bigcup _{m \le 2^n} A_{n,m}\) for all \(n\), and such that the sigma-algebra on \(\mathcal{Y}\) is the \(\sigma \)-algebra generated by the sets \(A_{n,m}\). Let \(\mathcal{F}\) be the filtration generated by the successive partitions, that is, \(\mathcal{F}_n = \sigma (\{ A_{n,m} \mid m \le 2^n\} )\).

TODO

This is the set of finite measures \(\{ Q' \mid Q' \le Q\} \). Indeed, if \(Q' \le Q\), then for all \(f\) with \(Q[f] \le 1\), \(Q'[f] \le Q[f] \le 1\). Conversely, if \(Q'\) is such that \((\forall f, \ Q[f] \le 1 \implies Q'[f] \le 1)\), then for any measurable set \(A\), we can take the measurable function \(f = Q(A)^{-1} \cdot \mathbb {I}_A\) (in which \(0^{-1} = \infty \)) to obtain \(Q'(A) \le Q(A)\).

Let \(\kappa \circ Q \in \kappa \circ \mathcal{Q}_{\operatorname{eff}}\) and \(f\) such that \(\forall Q' \in \mathcal{Q}, (\kappa \circ Q')[f] \le 1\). That is, \(\forall Q' \in \mathcal{Q}, Q'[x \mapsto \kappa (x)[f]] \le 1\). Then since \(Q \in \mathcal{Q}_{\operatorname{eff}}\), we have \(Q[x \mapsto \kappa (x)[f]] \le 1\). We have proved that \(\kappa \circ Q \in (\kappa \circ \mathcal{Q})_{\operatorname{eff}}\).

Since \(\mathcal{Q}_1 \subseteq \mathcal{Q}_2\) we have \(\mathcal{E}_{\mathcal{Q}_2} \subseteq \mathcal{E}_{\mathcal{Q}_1}\) (Lemma 2.2). Then

Since \(\operatorname{KL}\) is non-increasing in the second argument,

Since \(\{ Q\} _{\operatorname{eff}} = \{ Q' \in \mathcal{M}(\mathcal{X}) \mid Q' \le Q\} \) (Lemma 2.53), we obtained \(\operatorname{KL}(P, Q) = \inf _{Q' \in \{ Q\} _{\operatorname{eff}}} \operatorname{KL}(P, Q')\). By the duality of numeraire and \(\operatorname{KL}\) divergence (Theorem 2.56), we have

By Lemma 2.57, \(\operatorname{KL}(P, Q) = \sup _{f \in \mathcal{E}_{\{ Q\} }} P[\log f]\). Then by Lemma 2.24, we have

And that last term is \(\operatorname{KL}(\kappa \circ P, \kappa \circ Q)\), again by Lemma 2.57.