2.9

This exercise is asking us to prove Tchebychev's Inequality, stated as:

Suppose $f \geq 0$, and $f$ integrable. If $\alpha > 0$ and the set $E_\alpha$ is defined as: $$E_\alpha = \lbrace x | f(x) > \alpha \rbrace$$ Then: $$ m(E_\alpha) \leq \frac{1}{\alpha} \int f $$

Simply re-write the definition of our sets $E_\alpha$: $$E_\alpha = \lbrace x | \frac{f(x)}{\alpha} > 1 \rbrace $$ Next, observe that: $$m(E_\alpha) = \int_{E_\alpha} dx \hspace{0.25cm} \leq \int_{E_\alpha} \frac{f(x)}{\alpha} dx \hspace{0.25cm} \leq \frac{1}{\alpha} \int f(x) dx $$ ...as desired.