In my lecture we proved Jensen's inequality:
For any convex and measurable function $g : \mathbb{R} \rightarrow \mathbb{R}$ and random variable $x \in \mathbb{R}$ , it holds that: $$\mathbb{E}(g(x)) \geq g(\mathbb{E}(x)) $$ Here is our short proof: Linearize the function $g$ in $\mathbb{E}(x)$. Note that $g$ is measurable, so the expected value $\mathbb{E}$ doesn't make a problem. We know that $g$ is convex, so there are $s,t \in \mathbb{R} $ with: $g(x) \geq sx + t \ \forall x \in \mathbb{R}$ and $g(\mathbb{E}(x)) = s\mathbb{E}(x) + t $. Now $\mathbb{E}(g(x)) \geq \mathbb{E}(sx+t) = s\mathbb{E}(x)+ t =g(\mathbb{E}(x))$.
So I don't know how to prove why a convex function $g$ is measurable and I don't know how to prove why there are $s,t \in \mathbb{R}$ with $g(x) \geq sx + t \ \forall x \in \mathbb{R}$ and $g(\mathbb{E}(x)) = s\mathbb{E}(x) + t $. Here is my attempt:
For the first claim: If we want to check measurability of $g$ , we have to check if { $g < \alpha$ } $\in {B}(\mathbb{R}) $ ( Borel - $\sigma$ - algebra ). We know that $g$ is convex, so it holds that: $g(cx +(1-c)y) \leq cg(x) + (1-c)g(y) \ \forall x,y \in \mathbb{R} $ and $ \forall c \in [0,1] $.
If $g(x) \geq \alpha$ ,{ $g < \alpha$ } is the empty set $ \in {B}(\mathbb{R}) $ and if $g(x), g(y) < \alpha$ for some $x,y \in \mathbb{R}$ , we have $ g(cx +(1-c)y) \leq cg(x) + (1-c)g(y) < \alpha$ ? maybe this could help somehow? Moreover I read that the inverse image of a convex function is convex, and thus an interval (without proof). Maybe this could be useful, if we prove that? Like you see I really need help here.
For the second claim: For all $ x < u < y $, we have $u = cx + (1-c)y$ with $c = \frac{y-u}{y-x}$ Thus $ g(u) \leq cg(x) + (1-c)g(y) $, which implies $\frac{g(u)-g(x)}{u-x} \leq \frac{g(y)-g(u)}{y-u}$. Thus, taking $\sup_{x<u} \frac{g(u)-g(x)}{u-x} \leq s \leq \inf_{u<y} \frac{g(y)-g(u)}{y-u} $, we have that for all $x < y \in \mathbb{R}$, $g(x) \geq s(x-u) + g(u) = sx + (g(u)-su)$. For the second part of the second claim: I need your help again.