Does Jensen's inequality still hold in general finite measure space? 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
$$\psi\int_X f d\mu \le \int_X (\psi \circ f)d\mu$$
 A: 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.
A: I am very appreciated for the answer given by Nate Eldredge and Keba.
And I carefully reviewed proof process of Jensen's Inequality and found Why "Jensen's inequality Do Not hold in general finite measure space".
I write this thought down for anyone who has the same confusion.
In the original Proof:

Proof:
Since $\psi$ is convex, at each $x_0 \in \mathbb R$, there exist $a,b
 \in \mathbb R$ such that $\psi(x_0) = ax_0 + b$ and $\psi(x) \ge ax +
 b, \forall x \in \mathbb R$, (here, $y = ax + b$ defines a supporting
plane of the epigraph of $\psi$ at $x_0$). Let $x_0 = \int_X fdµ$,
then we have $$\psi(\int_Xf d\mu) = \psi(x_0) = ax_0+b=a\int_Xf\mu + b
 = \int(af+b)d\mu \le \int(\psi\circ f)d\mu$$, q. e. d.

When $\mu$ is a general finite measure, below equation do not hold:
$$a\int_Xf\mu + b  = \int(af+b)d\mu $$
Specifically, $$b  \neq \int b d\mu $$
In other words, below equations hold only when $\mu$ is a probability measure:
$$\int_X c\  d\mu = c , \ (c\  is\  constant)$$
$$E[E(x)] = E(x)$$
