Exercise 1.42 (Probability of a non-null set / expectation inequality)

The trick is to first prove the theorem assertion for simple functions. Hence, let $X=\sum <!--\; FUNCTION\; APPLICATION\; -->a_i{1}_{E_i}$ for non-zero $a_1,\dots ,a_n\in {ℝ}_n$ and measurable disjoint $E_1,\dots ,E_n$.

We prove the assertion by inducting on $n\in \mathbb{N}$.

• (induction base) Let $n=1$. We then have

  $$P(X\ne 0)=m(E_1)\ge m(E_1)=\frac{{(|a_1|m(E_1))}^2}{|a_1{\text{|}}^{2}m(E_1)}=\frac{{(\mathbf{E}|X|)}^2}{\mathbf{E}(|X{\text{|}}^2)}.$$

• (induction step) Now suppose inductively that we have proven the theorem assertion for some $n\in N$. We now analyze the case for $n\text{++}$. The assertion

  $$\mathbf{P}(X\ne 0)\ge \frac{{(\mathbf{E}|X|)}^2}{\mathbf{E}(|X{\text{|}}^2)}$$

  is equivalent to

  $$\mathbf{P}(X\ne 0)\cdot \mathbf{E}(|X{\text{|}}^2)\ge {(\mathbf{E}|X|)}^2.$$

  In other words, we must demonstrate that

  $$\left(\sum _{j=1}^{n+1}m(E_j)\right)\cdot \left(\sum _{i=1}^{n+1}|a_i{\text{|}}^{2}m(E_i)\right)\ge {\left(\sum _{i=1}^{n+1}\left|a_i\right|m(E_i)\right)}^2.$$

  We have

  $$\begin{array}{llll}\hfill & \sum _{j=1}^{n+1}\sum _{i=1}^{n+1}|a_i{\text{|}}^{2}m(E_i)m(E_j)& \hfill & \\ \hfill & =\sum _{j=1}^n\sum _{i=1}^{n+1}|a_i{\text{|}}^{2}m(E_i)m(E_j)+\sum _{i=1}^{n+1}|a_i{\text{|}}^{2}m(E_i)m(E_{n+1})& \hfill & \\ \hfill & =\sum _{j=1}^n\sum _{i=1}^n|a_i{\text{|}}^{2}m(E_i)m(E_j)+\sum _{j=1}^n|a_{n+1}{\text{|}}^{2}m(E_{n+1})m(E_j)+\sum _{i=1}^{n+1}|a_i{\text{|}}^{2}m(E_i)m(E_{n+1})& \hfill & \\ \hfill & =\sum _{j=1}^n\sum _{i=1}^n|a_i{\text{|}}^{2}m(E_i)m(E_j)+\sum _{j=1}^n|a_{n+1}{\text{|}}^{2}m(E_{n+1})m(E_j)+\sum _{i=1}^n|a_i{\text{|}}^{2}m(E_i)m(E_{n+1})+|a_{n+1}{\text{|}}^{2}m{(E_{n+1})}^2& \hfill & \\ \hfill & =\sum _{j=1}^n\sum _{i=1}^n|a_i{\text{|}}^{2}m(E_i)m(E_j)+\sum _{i=1}^{n}m(E_i)m(E_{n+1})\left[|a_i{\text{|}}^2+|a_{n+1}{\text{|}}^2\right]+|a_{n+1}{\text{|}}^{2}m{(E_{n+1})}^2& \hfill & \\ \hfill & \stackrel{\mathit{IH}}{\text{≥}}{\left[\sum _{i=1}^n\left|a_i\right|m(E_i)\right]}^2+2\cdot \sum _{i=1}^{n}m(E_i)m(E_{n+1})\left[\left|a_i\right|\cdot \left|a_{n+1}\right|\right]+{\left[\left|a_{n+1}\right|m(E_{n+1})\right]}^2& \hfill & \\ \hfill & ={\left(\sum _{i=1}^{n+1}\left|a_i\right|m(E_i)\right)}^2.& \hfill & \end{array}$$

  This completes the induction.

Now we proceed with the general case. Hence, suppose that $X$ is measurable double integrable and let ${(Y_n)}_{n\in \mathbb{N}}$ be a sequence of non-decreasing simple functions converging to $X$. We have already proven that

Let’s look at the asymptotic behaviour of this sequence:

On the one hand, by the monotone convergence theorem (Theorem 18) we have

On the other hand, the sets $\{Y_n\ne 0\}\supseteq \{Y_{n+1}\ne 0\}$ are decreasing; thus, by downwards monotone convergence (Exercise 0.23) we obtain

as desired.