Prove that if $m$ is a finite measure, then $m\left( \bigcap^{k}_{i=1}A_{i} \right)\leq \sqrt[k]{\prod^{k}_{i=1}m(A_{i} )}$

Question

I want to prove that if $m$ is a finite measure, then

$$ m\left( \bigcap^{k}_{i=1}A_{i} \right)\leq \sqrt[k]{\prod^{k}_{i=1}m(A_{i} )}$$

MY TRAIL

For $n=1,m\left( A_{1} \right)\leq m(A_{1} )$

We assume it is true for some $k\geq 0,$ i.e. $$ m\left( \bigcap^{k}_{i=1}A_{i} \right)\leq \sqrt[k]{\prod^{k}_{i=1}m(A_{i} )}$$ So, \begin{align} m\left( \bigcap^{k+1}_{i=1}A_{i} \right)&\leq \sqrt[k+1]{\prod^{k+1}_{i=1}m(A_{i} )}\\&= \sqrt[k+1]{\prod^{k}_{i=1}m(A_{i} )\times m(A_{k+1} )}\end{align}

I got stuck at this point. I know that I have not used the hypothesis that $m$ is finite. Can anyone show me how to continue?

Answer 1

The measure of $\bigcap^{k}_{i=1}A_{i} $ is at most the measure of the $A_i$ with the smallest measure since it is a subset of this $A_i$. Then use that the geometric mean of $k$ nonnegative numbers is at least the minimum of these numbers.