Show: $\int_{[a,b]}{f d\lambda^d}=0 \:\:\:\:\forall a, b \in \Bbb R ^d \:\:\mathrm{ with }\:\: a \le b\Rightarrow f=0 \:\:\mathrm{ a.e.}$

Question

Let $f: \Bbb R^d \to \Bbb R$ be integrable. I have to show: $$\int_{[a,b]}{f d\lambda^d}=0 \:\:\:\:\forall a, b \in \Bbb R ^d \:\:\mathrm{ with }\:\: a \le b\Rightarrow f=0 \:\:\mathrm{ a.e.}$$

The hint was given to first show that the system of all measurable sets with $\int_A {f d\lambda^d}=0$ (i called it $\mathcal A$) form a $\sigma$-algebra. (I've shown that unions and the empty set are in $\mathcal A$ but I'm not quite sure about the step that for $A\in \mathcal A$, $A^c \in \mathcal A$ as well)

Any tipps or ideas on how to show the above relation? Thanks in advance!

Answer 1

Hint: Since $f$ is integrable, we know by dominated convergence that

$$\int_{\mathbb{R}^d} f = \lim \limits_{n \to \infty} \int_{[-n, n]^d} f = \lim \limits_{n \to \infty} 0 = 0$$