# 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 $$\psi\int_X f d\mu \le \int_X (\psi \circ f)d\mu$$

• I think it holds, but not very sure. Commented Jul 20, 2020 at 18:49
• Take $X = [0, a] \subset \mathbb R$, $f \equiv 1$ and $\varphi(s) = s^p$ for some $p > 1$ and look what happens.
– Keba
Commented Jul 20, 2020 at 19:05
• Thanks, Keba. That is very helpful! Commented Jul 21, 2020 at 8:26

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.

• Thanks very much, Nate Eldredge. That is very helpful! Commented Jul 21, 2020 at 8:28

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: $$\int_X c\ d\mu = c , \ (c\ is\ constant)$$ $$E[E(x)] = E(x)$$