TITLE: Limit measure is zero?
QUESTION [6 upvotes]: Let $E\subset \mathbb{R}$ be a Borel set and suppose that $(f_k)_{k\in\mathbb{N}}$ is a sequence of Lebesgue measurable functions defined on $E$.
Assume that $\lim_{k\to+\infty}f_k(x)=f(x)$ for almost every $x$ and that there exists a Lebesgue integrable function $g$ such that $|f_k(x)|\leq g(x)$ for almost every $x\in E$ and every $k\in\mathbb{N}$. Fix $\varepsilon >0$ and define $A_k^\varepsilon=\{x\in E : |f_k(x)-f(x)|\geq\varepsilon\}$. Is it true that
$$
\lim_{j\to+\infty}\left|\bigcup_{k\geq j} A_k^\varepsilon \right| =0 \quad ?
$$
($|\cdot|$ is the Lebesgue measure)
The only thing that I noticed is that $|A_k^\varepsilon|\xrightarrow[k\to+\infty]{}0$, as a consequence of Lebesgue's dominated convergence theorem.
Does anyone know if this is true or not (and why?)?

REPLY [4 votes]: True. For every $j \in \{1,2,\dots\}$ define $B_j := \bigcup_{k\geq j}A_k^\varepsilon$. Then $B_1 \supseteq B_2 \supseteq B_2\supseteq \cdots$, and $\lambda\big(\bigcap_{j=1}^\infty B_j\big) = \lambda\big(|f_k(x)-f(x)|\geq\varepsilon\quad i.o.\big) = 0$, since $\lim_{k\rightarrow\infty}f_k(x)=f(x)$ almost everywhere. Hence, if we can show that $\lambda(B_1) < \infty$, then by the continuity of measures $\lim_{j\rightarrow\infty}\lambda(B_j) = 0$. In fact, $\lambda(B_1) \leq \lambda\big(g\geq\frac{\varepsilon}{2}\big)<\infty$: the first inequality is due to the fact that if $x \in B_1$, then $|f_k(x)-f(x)|\geq\varepsilon$ for some $k\in\{1,2,\dots\}$, and almost everywhere $|f_k(x)|, |f(x)| \leq g(x)$; and the second inequality is due to the fact that $g$ is integrable.