Show $\int_Af(x)\,dx=0$ for every measurable subset $A$ of $[0,1]$.

Question

Let $f$ be Lebesgue-integrable on $[0,1]$. Suppose $\int_a^bf(x)\,dx=0$ for all $0\leq a\leq b\leq 1$. Show $\int_Af(x)\,dx=0$ for every measurable subset $A$ of $[0,1]$.

*Let $A$ be a measurable subset of $[0,1]$. Then $A$ can be written as the union of disjoint, countable? intervals. Since $\int_a^bf(x)\,dx=0$ for all $0\leq a\leq b\leq 1$, each integral of $f$ over each interval is $0$ so $\int_Af(x)\,dx=0$.

I'm not sure if I did it right...

Answer 1

The Lebesgue measure is regular, and in particular outer regular. Therefore there exists a sequence $(O_n)$ of open measurable sets containing $A$ such that $\lambda(A)=\inf \lambda(O_n)$.

Then $\displaystyle \int_A f(x)\, \mathrm{d}x=\int_{O_n} f(x)\, \mathrm{d}x - \int_{O_n - A} f(x)\, \mathrm{d}x$.

Since $O_n$ is an open set of $\mathbb{R}$, it is a countable union of open intervals and $\int_{O_n} f(x)\, \mathrm{d}x=0$. By dominated convergence you can show that $\int_{O_n - A} f(x)\, \mathrm{d}x \to 0$: use the sequence of functions $f1_{O_n-A}$ dominated by $|f|$.

Answer 2

This is a standard application of Dynkin's $\pi$-$\lambda$ Theorem.

Define $\mathcal{P}=\{(a,b]\mid0\leq a<b\leq1\}\cup\{\emptyset\}$ and $\mathcal{L}=\{A\in\mathcal{B}([0,1])\mid\int_{A}f(x)\,dx=0\}$. It is routine to verify that $\mathcal{P}$ is a $\pi$-class and $\mathcal{L}$ is a $\lambda$-class. That is:

• For any $A,B\in\mathcal{P}$, we have $A\cap\mathcal{B\in\mathcal{P}}$,
• $\emptyset\in\mathcal{L}$,
• $A^{c}\in\mathcal{L}$ whenever $A\in\mathcal{L}$, where the complement is taken with respect to $[0,1]$,
• $\cup_{i=1}^{\infty}A_{i}\in\mathcal{L}$ whenever $A_{i}\in\mathcal{L}$ and $A_{1},A_{2}\ldots$ are pairwisely disjoint.

By the given condition, we have $\mathcal{P}\subseteq\mathcal{L}$. Therefore, by Dynkin's $\pi$-$\lambda$ Theorem, we have $\sigma(\mathcal{P})\subseteq\mathcal{L}$. Note that $\sigma(\mathcal{P})=\mathcal{B}([0,1])$. Therefore, $\mathcal{L=B}([0,1])$. Q.E.D.