# Jensen's Inequality Proof for Conditional Expectation (Durrett)

I have some doubts as I read over Durrett's proof on Jensen's inequality for conditional expectation.

The statement is that if $$\varphi$$ is convex and $$E|X|, E|\varphi(X)|<\infty$$, then $$\varphi(E(X|\mathcal{F})) \leq E(\varphi(X)|\mathcal{F})$$.

In his proof he worked through the case where $$\varphi$$ is non-linear (as the linear case is trivial). I also know that any convex function is the supremum of some collection of affine functions.

However, what I do not understand is that he let $$S = \{(a, b): a,b\in \mathbb{Q}, ax+b \leq \varphi(x), \forall x\}$$ and let $$\varphi(x) = \sup\{ax + b:(a, b) \in S\}$$. He also said that there is an exceptional set for each a, b, which violates the Jensen's inequality, so we have to take the $$\sup$$ over a countable set (which is why he defined $$S$$ in such a way I suppose).

My questions are:

1. Why is this definition of $$\varphi$$ (by using rational a and b) sufficient to cover all possible non-linear convex functions?

2. I understand that there might be $$a$$ and $$b$$ such that $$E(\varphi(X) | \mathcal{F}) < aE(X | \mathcal{F}) + b$$, but such an event will have 0 measure (which is why he stressed A.S.). But why is taking $$\sup$$ over a countable set solves the problem that there is an exceptional set for each a, b?

Here is the screenshot of the theorem and the proof; problematic statements are highlighted in blue:

Thank you very much.

1. It is essentially because, for each interior point $$x_0$$ of the domain of $$\varphi$$, we can find $$m$$ such that $$\varphi(x) \geq \varphi(x_0) + m(x - x_0).$$ Any such line is called a supporting line. Now, each supporting line can be approximated by the family of lines $$y=ax+b$$ parametrized by $$S$$.
(Actually, the claim, $$\varphi(x) = \sup\{ax+b : (a, b) \in S \}$$, may fail at points of discontinuity of $$\varphi$$. However, such points are necessarily the endpoints of the domain of $$\varphi$$, and such case can be treated separately.)
If $$X_n\leq Y$$ a.s. for each $$n$$, then $$\sup_n X_n\leq Y$$ a.s.
This is because $$\mathbb{P}(\sup X_n > Y) = \mathbb{P}\bigl(\cup_n \{X_n > Y\}\bigr) \leq \sum_n \mathbb{P}(X_n > Y) = 0.$$ Now the highlighted part follows from exactly the same reason.