# Jensen's inequality; what's the need for the probability measure?

Jensen's inequality states that if: $\mu$ is a probability measure on $\Omega$, $f$ is integrable function (on $\Omega$) and $\phi$ is convex on the range of f then:

$\phi \left( \int_{\Omega}g(x)\,d\mu \right) \le \int_{\Omega}\phi \left( g(x) \right)d\mu$

I wonder why $\mu \left( \Omega \right)$ has to be finite and equal to one? Why can't one let $\Omega$ = $\mathbb{R}$ for example?

If you consider $\mu$ to be an arbitrary finite measure and consider normalizing it to $\nu$, then applying the ordinary Jensen inequality to $\nu$, you get another version:
$$\phi \left ( \frac{1}{\mu(\Omega)} \int_\Omega g d \mu \right ) \leq \frac{1}{\mu(\Omega)} \int \phi \circ g d \mu.$$
The proof itself completely breaks down when $\mu(\Omega)=+\infty$, though.
You just have to consider $\mu$ be a finite measure, i.e. $\mu(\Omega)<\infty$, because if not you can´t apply the convexity of $\phi$. First let observe that without loss of generality we can consider $\mu(\Omega)=1$, because if $\mu(\Omega)<\infty$ then tou can take another measure that is normalized i.e. $d\nu(x)=\frac{d\mu(x)}{\mu(\Omega)}$. Also if $\phi$ is a convex function you have $\phi(y)\geq ay+b$ for $a,b$ constants, and if you fix $y_{0}$ you also have that $\phi(y_{0})=ay_{0}+b$, then $$\phi\left(\int_{\Omega}f(x)d\mu(x)\right)=a\int_{\Omega}f(x)d\mu(x)+b,$$ now using that $\mu(\Omega)=1$ you have $$a\int_{\Omega}f(x)d\mu(x)+b=\int_{\Omega}(af(x)+b)d\mu(x)\leq \int_{\Omega}\phi(f(x))d\mu(x).$$ So as you see if you don´t consider a finite measure, you can´t do this trick because you need to have that the integral of a constant is finite.
• Just to be sure I get it: $\int_{\Omega}(af(x)+b)\,d\mu(x)$ =$a \int_{\Omega}f(x)d\mu(x) + \int_{\Omega}b\,d\mu(x)=a \int_{\Omega}f(x)d\mu(x) + \mu(\Omega)b.$ And since $\mu(\Omega)=1$ this is equal to $a \int_{\Omega}f(x)d\mu(x) +b$. This step won't work if $\mu(\Omega)\neq 1$? – user202542 Mar 5 '16 at 19:43