# 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:

$$\phi\left(\int_\Omega g(x)\,\mathrm d\mu\right)\leq\int_\Omega\phi\left[g(x)\right]\,\mathrm d\mu.$$

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:

$$\phi \left ( \frac{1}{\mu(\Omega)} \int_\Omega g d \mu \right ) \leq \frac{1}{\mu(\Omega)} \int \phi \circ g d \mu.$$

In either case, it says applying a convex function to the average gives a smaller result than averaging the convex function itself.

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 $\mu(\Omega)\neq 1$? Commented Mar 5, 2016 at 19:43

A quick collection of counterexamples why the Jensen equality in it's usual form fails in other cases:

General counterexample for finite measures $$\mu(X) = c \in [0,\infty) \setminus\{1\}$$:

Every affine linear function $$\phi(x) = ax+b$$ is convex. Suppose Jensen's equality holds, then $$a \mu(X) + b = a\left(\int_X 1\ d\mu \right) + b \leq \int_X a+b\ d\mu = (a+b)\mu(X).$$ Thus $$b(1- \mu(X)) \leq 0$$ for all $$b\in\Bbb R$$, as $$\phi$$ can be an arbitrary affine linear map. But clearly we can choose some $$b\in\Bbb R$$ such that this fails.

Specific example for infinite measure: Take $$f(x)=x^{-1}, \phi(x)=x^2$$ on $$[1,\infty[$$ with the Lebesgue measure. Then $$\phi\left(\int_{[1,\infty[} x^{-1}\ d\lambda(x)\right) = \infty \not\leq \int_{[1,\infty[} x^{-2}\ d\lambda(x) < \infty.$$