I am trying to understand a proof in Stein and Shakarchi of the following Lemma, but am having some difficulty.
Let $f$ be a bounded function supported on a set $E$ of finite measure. If $\{\varphi_n\}_{n=1}^\infty$ is any sequence of simple functions bounded by $M$, supported on $E$, and with $\varphi_n(x)\rightarrow f(x)$ for a.e. $x$, then the limit $\lim_{n\rightarrow\infty}\int\varphi_n$ exists.
The proof begins as follows: Since the measure of $E$ is finite, given $\epsilon>0$ Ergov's theorem guarantees the existence of a (closed) measurable subset $A_\epsilon$ of $E$ such that $m(E-A_\epsilon)\leq \epsilon$, and $\varphi_n\rightarrow f$ uniformly on $A_\epsilon$. Therefore, setting $I_n=\int\varphi_n$ we have that $$|I_n-I_m|\leq\int_E|\varphi_n(x)-\varphi_m(x)|dx=\int_{A_\epsilon}|\varphi_n(x)-\varphi_m(x)|dx+\int_{E-A_\epsilon}|\varphi_n(x)-\varphi_m(x)|dx\leq\int_{A_\epsilon}|\varphi_n(x)-\varphi_m(x)|dx+2Mm(E-A_\epsilon)$$.
This last inequality is what I'm confused about. Why is it the case that $\int_{E-A_\epsilon}|\varphi_n(x)-\varphi_m(x)|dx\leq 2Mm(E-A_\epsilon)$?
Best Answer
It comes from the boundedness of the $\varphi_n$:
$$|\varphi_n - \varphi_m| \le |\varphi_n| + |\varphi_m| \le 2M.$$
And so $$ \int_{E - A_\epsilon} |\varphi_n - \varphi_m|d\mu \le \int_{E- A_\epsilon} 2M d\mu = 2M \,m(E- A_\epsilon). $$