# How to prove that $m\left(\lbrace f>\alpha\rbrace\right)$ is a right-continuous function of $\alpha$?

Let $f$ be a Lebesgue measurable function that is finite almost everywhere on $\left[a,b\right]$. Prove that $m\left(\lbrace f>\alpha\rbrace\right)$ is a right-continuous function of $\alpha$, and that $m\left(\lbrace f\geq\alpha\rbrace\right)$ a left-continuous function of $\alpha$.
I am really curious about how ''>'' and ''$\geq$'' make such a difference.

• To see why it makes a difference, compute the two functions for something simple like $f=1_{(0,1)}$. Nov 1 '15 at 6:31
• @PhoemueX Thanks for your example. But I still have no idea how this difference is reflected in the proof. Let $\epsilon>0$ be given. We need to show that there is $\delta>0$ such that $\vert m\left(\lbrace f>\alpha\rbrace\right)-m\left(\lbrace f>\alpha^{\prime}\rbrace\right)\vert=m\left(\lbrace \alpha\leq f<\alpha^{\prime}\rbrace\right)<\epsilon$ for any $\alpha^{\prime}\in\left(\alpha,\alpha+\delta\right)$, taking the first half for instance. So what follows? There must be something to do with "a.e. finite" I suppose. THX! Nov 1 '15 at 7:22
• Forget about the proof, try to compute that two function for the constant function $f(x)=1$ first, as suggested by PhoemueX.
– user99914
Nov 1 '15 at 7:34

1. $$A_1\subset A_2\subset A_3……$$ and $$A=\cup_{i\in N} A_i$$ then $$m(A)=\lim_{i\rightarrow\infty} m(A_i)$$
2. $$A_1\supset A_2\supset A_3……$$ and $$A=\cap_{i\in N} A_i$$ and $$m(A_1)<\infty$$ then $$m(A)=\lim_{i\rightarrow\infty} m(A_i)$$
Now define $$\phi(\alpha)=m(\{f>\alpha\})$$, we have $$\phi(\alpha)=m(\{f>\alpha\})=m(\cup_{n\in N} \{f>\alpha+\frac{1}{n}\})=\lim_{n\rightarrow \infty} m(\{f>\alpha+\frac{1}{n}\})=\lim_{n\rightarrow \infty}\phi(\alpha+\frac{1}{n})$$ by property 1. I'll left the proof of $$\phi(\alpha)=\lim_{n\rightarrow \infty}\phi(\alpha+\frac{1}{n})$$ implies right-continuous definition to you.
For the other one, observe $$\{f\geq \alpha\}=\cap_{n\in N}\{f\geq\alpha-\frac{1}{n}\}$$, use property 2 and the similar approach.