I'm trying to understand an argument my prof made:
Given $h:\mathbb R \rightarrow \mathbb R$ convex, we look at $E[h(X)]$. The expected value exists because $h(X)$ is lower-bounded by [some] $l(X)$ that is integrable.
I'm not certain why this lower bound implies that the expected value exists. I know, of course, Lebesgue's theorem about dominated convergence but that refers to an upper bound.
So I tried to search for a different reason behind it and I came up with this: We have $E(h(X)^-) \leq E(l(X)^-) < \infty$ since $l(X) \leq h(X)$. Our lecture then implies that $E(h(X))$ exists already.
But couldn't we just use that argument in generalization and we get that whenever an $X$ is lower bounded by some integrable $Y$ random variable, then $E(X)$ exists.
Is that correct?
Assuming $X$ to be integrable: By convexity, $h$ is bounded below by affine $l$; and $l(X)$ is integrable because $X$ is so. In particular, $E[l(X)^-]<\infty$. (Where $b^-:=\max(-b,0)$.) Therefore, because $h(x)^-\le l(x)^-$ for all $x$, $E[h(X)^-]\le E[l(X)^-]<\infty$. Consequently, $E[h(X)]$ is well defined as $E[h(X)^+]-E[h(X)^-]$, although $E[h(X)]$ might take the value $+\infty$.