Let $A$ be a measurable subset of $[0,2]$ and define $f:\mathbb{R}\to\mathbb{R}$ by letting $f(x) = \mu((-\infty,x]\cap A)$ for every $x\in\mathbb{R}$.

(a) Show that $f$ is absolutely continuous on $\mathbb{R}$, calculate $f'$ and $\int_0^3f'(x)dx$

(b) Show that for every $0<b<\mu(A)$ there exists $x_0\in\mathbb{R}$ such that $b=\mu((-\infty,x_0]\cap A)$

For (a), I was able to show that $f$ is absolutely continuous by applying the definition that for every $\varepsilon>0$ there is a $\delta>0$ such that for every finite disjoint collection $\{(a_k,b_k) \}_{k=1}^n$ of open intervals then $\sum b_k-a_k < \delta \Rightarrow \sum|f(b_k)-f(a_k)|<\varepsilon$. Taking $\delta=\varepsilon$, it's pretty easy to see that this definition is satisfied.

I also know that since $f$ is absolutely continuous, then $\int_a^bf'(x)dx$=f(b)-f(a), so $\int_0^3f'(x)dx= f(3)-f(0) = \mu(A)$.

I'm having trouble actually calculating what $f'$ is though. I tried using the difference quotient, and I simplify it down to $$f'=\lim_{h\to0}\frac{\mu((t,t+h]\cap A)}{h} $$ I'm not sure where to proceed from here though, if this is even the right way to go. Any help would be much appreciated.

I think part (b) is simply the Intermediate Value Theorem, but if it's not I would appreciate a hint.


1 Answer 1


Note that $$f (x)=\int_0^x 1_A (t)\,d\mu. $$ Then $$\frac {f (x+h)-f (x)}h=\frac1h\,\int_x^{x+h}1_A (t)\,d\mu (t)\xrightarrow[h\to0]{} 1_A (x) $$ a.e., by the differentiation theorem.

Part b is indeed the intermediate value theorem.


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