I cannot work out this problem even though it seems not that difficult. Could anyone kindly give me any hint? Thanks!
If $f(x)$ is measurable on $E \subset \mathbb R$, then $$ \varphi (t)=m\big(\{x \in E : f(x) \lt t\}\big) $$ is a left continuous function on $\mathbb R$.
I first picked a sequence $\{t_n\}$ converging to $t$.
But I cannot prove that $$ \lim_{t_n \nearrow t} \,m\big(\{x \in E : f(x) \lt t_n\}\big)=m\big(\{x \in E : f(x) \lt t\}\big). $$
On
$g$ is left continuous at $t_0$ means: If $t_n\to t_0$ and $t_n<t_0$, then $g(t_n)\to g(t_0)$. In fact, we can assume that $\{t_n\}$ is strictly increasing.
In our case, $$ \{x\in E: f(x)<t_0\}=\{x\in E: f(x)<t_1\}\cup\{x\in E: t_1\le f(x)<t_2\}\cup\cdots\cup \{x\in E: t_n\le f(x)<t_{n+1}\}\cup\cdots=\{x\in E: f(x)<t_1\}\cup\bigcup_{n=1}^\infty \{x\in E: t_n\le f(x)<t_{n+1}\} $$ and all the sets $$ \{x\in E: f(x)<t_1\},\, \{x\in E: t_n\le f(x)<t_{n+1}\}, \quad n\in\mathbb N, $$ are measurable and disjoint. Thus
$$ g(t_0)=m\left(\{x\in E: f(x)<t_0\}\right)=m\left(\{x\in E: f(x)<t_1\}\right) +\sum_{n=1}^\infty m\left(\{x\in E: t_n \le f(x)<t_{n+1}\}\right) \\ =m\left(\{x\in E: f(x)<t_1\}\right) +\lim_{n\to\infty}\sum_{k=1}^n m\left(\{x\in E: t_k\le f(x)<t_{k+1}\}\right)=\lim_{n\to\infty} m\left(\{x\in E: f(x)<t_{n+1}\}\right)=\lim_{n\in\infty} g(t_{n+1}). $$
For any $t$, let $S_t:=\{x\in E|f(x)<t\}$, and note that whenever $(t_n)_{n=1}^\infty$ is an increasing sequence that converges to $t$ we have $S_t=\bigcup_nS_{t_n}$, thus $$m(S_t)=m\left(\bigcup_nS_{t_n}\right)=\lim_{n\to\infty}m(S_{t_n}),$$and left continuity follows.