I got some useful information from this Question: Jensen's inequality in measure theory
Theorem 3.1 Jensen's Inequality
Let $(X,\mathcal{M},\mu)$ be a probability space (a measure space
with $\mu(X) = 1$ ), $f: X \to \mathbb R \in L^1(X, \mu)$, and $\psi:\mathbb R \to \mathbb R $ be a convex function, then
$$\psi\int_X f d\mu \le \int_X (\psi \circ f)d\mu$$
And that question asked whether Jensen's inequality still hold in general finite measure space ?
A nice man d.k.o. answered:
Yes. In this case for convex $\varphi$ :$$\varphi\left(\frac{1}{\mu(X)}\int fd\mu\right)\le \frac{1}{\mu(X)}\int \varphi\circ fd\mu$$
However, this result is basically rescale $\mu$ to a probability measure.
So whether the following proposition hold?
Let $(X,\mathcal{M},\mu)$ be a general measure space, and $\mu(X) < \infty $,
$f: X \to \mathbb R \in L^1(X, \mu)$, and $\psi:\mathbb R \to \mathbb R $ be a convex function, then
$$\psi\int_X f d\mu \le \int_X (\psi \circ f)d\mu$$
Best Answer
No. Indeed, Jensen's inequality in its basic form only holds if $\mu$ is a probability measure. Setting $f=1$ shows that we have $\psi(\mu(X)) \le \psi(1) \mu(X)$ for every convex function $\psi$. If $\mu(X) \ne 1$ then we could take $\psi$ to be a linear function with $\psi(1) = 0$ and $\psi(\mu(X)) > 0$, yielding a contradiction.