这是一份liverpool利物浦大学MATH365的成功案例

Now for each $n \geq 1$, let
Then, $\mu\left(A_{n}\right)<\frac{1}{2^{n}}$ and hence $$ \sum_{n=1}^{\infty} \mu\left(A_{n}\right)<\infty $$ By the MCT, this implies that $\int_{[a, b]}\left(\sum_{n=1}^{\infty} I_{A_{n}}\right) d \mu<\infty$ and hence $$ \sum_{n=1}^{\infty} I_{A_{n}}<\infty \text { a.e. } \mu \text {. } $$ Thus $h_{n} \rightarrow f_{K}$ a.e. $\mu$ on $[a, b]$. By Egorov’s theorem for any $\epsilon>0$, there is a set $A_{e} \in \mathcal{B}([a, b])$ such that
$$
\mu\left(A_{\epsilon}^{c}\right)<\epsilon / 2 \text { and } h_{n} \rightarrow f_{K} \text { uniformly on } A_{\tau} \text {. }
$$

MATH365 COURSE NOTES ：

Let $(\Omega, \mathcal{F}, \mu)$ be a measure space and let $f: \Omega \rightarrow[0, \infty]$ be such that it admits two representations
$$
f=\sum_{i=1}^{k} c_{i} I_{A_{i}} \text { and } f=\sum_{j=1}^{\ell} d_{j} I_{B_{j}},
$$
where $c_{i}, d_{j} \in[0, \infty]$, and $A_{i}$ and $B_{j} \in \mathcal{F}$ for all $i, j$. Show that
$$
\sum_{i=1}^{k} c_{i} \mu\left(A_{i}\right)=\sum_{j=1}^{\ell} d_{j} \mu\left(B_{j}\right) .
$$