Right-continuity of the function $g(t)=\mu(\{x\in A:f(x)>t\})$ Let $(X,M,\mu)$ be a measure space, $A\in M$ with $\mu(A)<\infty$ and $f:A\to \mathbb{R}$ be a measurable function on $A$. Show that the function $g(t)=\mu(\{x\in A:f(x)>t\})$ is non-increasing on $\mathbb{R}$ and right continuous in each point.
My approach: Let $t_1<t_2$ then $$\{x\in A: f(x)>t_1\}=\{x\in A:f(x)>t_2\}\sqcup \{x\in A: t_1<f(x)\leq t_2\}$$ and since each of those sets are measurable then by $\sigma$-additivity of $\mu$: $$g(t_1)=g(t_2)+\mu(\{x\in A: t_1<f(x)\leq t_2\})$$ and hence $g(t_1)\geq g(t_2)$.
Let's show that $g(t)$ is right-continuous on $\mathbb{R}$. Let's fix $t_0\in 
\mathbb{R}$ and take the sequence $t_n>t_0$ such that $t_n\to t_0$. Suppose that $E_n:=\{x\in A:f(x)>t_n\}$ for $n\geq 0$. Then it follows that $E_0=\bigcup \limits_{n=1}^{\infty} E_n$. If I can show that $\mu(E_0)=\lim \limits_{n\to \infty}\mu(E_n)$ then I am done.
But I cannot show this. I was trying to apply the continuity of measure $\mu$ but in this case $E_n$ is not a nested sequence.
Please help me to finish the solution.
 A: It is sufficient to deal with non increasing sequences.
By $t_n \downarrow t$ I mean a sequence $t_n \to t$ such that $t_n \ge t$ for all $n$.
First note that if $t_n \downarrow t$ then there is a non increasing subsequence $t_{n_k} \downarrow t$. (To see this, note that either $t_n = t $ infinitely often or $t_n >t$ infinitely often. In the latter case,
start with some $t_{n_1}>t$ and select the next element ${1 \over 2}t_{n_1} > t_{n_2}>t$, etc, etc.)
Then we have $g$ is continuous from the right at $t$ iff $g(t_n) \to g(t)$ for all non increasing sequences $t_n \downarrow t$.
To see this:
If $g$ is continuous from the right then it is clear that if
$t_n \downarrow t$ then $g(t_n) \to g(t)$.
Now suppose $g(t_n) \to g(t)$ for all non increasing sequences $t_n \downarrow t$. To prove by contradiction, suppose
$g$ is not continuous from the right at $t$. Then there is some $\epsilon>0$ and a sequence
$t_n \downarrow t$ such that $|g(t)-g(t_n)| \ge \epsilon$ for all $n$.
Now select a non increasing subsequence as above to get a contradiction.
Note: Extracting a non increasing subsequence from $t_n \downarrow t$.
If $t_n = t$ infinitely often then choose this subsequence.
Otherwise $t_n=t $ only happens for a finite number of terms
and after some $N$ we have $t_n >t$ for $n \ge N$.
Then, for any $n \ge N$ we can find some $n' >n$ such that
${1\over 2}t_n > t_{n'} > t$. Repeating this procedure we
can create a strictly decreasing subsequence.
