Counterexample to Jensen's Inequality when the convex function admits values in the extended real set

Question

Let $(X,A,\mu)$ be a set, a $\sigma$-algebra and a measure. Suppose that $\mu(X) = 1$. Let $u : X \rightarrow {\mathbb{R}}$ and $f : \mathbb{R} \rightarrow \mathbb{R} \, \cup \, \{+\infty \} $ be an integrable function and a convex, lower semi-continous function respectively. Then $$\int_{X} f \circ u \, d\mu \geq f \left( \int_{X} u \, d\mu \right)$$ This is the version of Jensen's Inequality I'm working with (note that convex functions with values in $\mathbb{R} \cup {+\infty}$ are not automatically continous and that LSC is required in order to have the lower affine approximants required to prove the theorem). I'd like to find a counterexample to this inequality when $f$ is convex but not necessarily lower semi-continous. I expected it to be quite easy to find, but the fact that $E = \{ x \in \mathbb{R} | f(x) < +\infty \}$ has to be an interval and that $u(x)$ has to be in $E$ for almost every $x \in X$ for the left-hand side of the equation not to be $+\infty$ is kinda annoying!

Answer 1

It seems to me the hypothesis of lower semi-continuity is not needed.

The left side of your inequality is well defined, though perhaps equal to $+\infty$. In this case the inequality holds trivially. If $\int_X f\circ u \,d\mu<\infty$, then $\mu(\{x\in X: u(x)\in\{f=+\infty\}\})=0$. Let $G$ denote $\{x\in X: u(x)\in\{f<+\infty\}\}$. Let $g$ be $f$ redefined (if necessary) at the finite endpoints of $\{f<+\infty\}$ so as to be lsc; that is, if $b$ is such an endpoint then $g(b):=\liminf_{t\to b} f(t)$. Then $$ \int_X f\circ u \, d\mu=\int_G f\circ u \, d\mu=\int_G g\circ u \, d\mu\ge g\left(\int_G u \, d\mu\right)=g\left(\int_X u \, d\mu\right)=f\left(\int_X u \, d\mu\right). $$ Here the inequality follows from the lsc form of Jensen that you noted, and the final equality is true because $\int_X u \, d\mu\in \{f<+\infty\}$.