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If a (Lebesgue) measurable function $f: ℝ→ℝ$ satisfies $∫_a^b f(x)dx = 0$ for any $a, b ∈ ℝ$, does $f(x)=0, a.e.$?

I understand that $f(x)=0$ follows if $f$ is continuous, but I was not sure if the same is true for a measurable function.

Edit:

Almost same question was here: Prove that the Lebesgue integral of $f\chi$ equals to $0$ indicates that $f=0$ a.e.

(I missed it, thank you for letting me know!)

Jaborandi Kakapo
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    Thanks for the info, José Carlos Santos. But it looks like a different problem to me. The problem in your link is about a function whose range is [0, ∞), but in my question, it's important that the range of f is ℝ, not [0, ∞).

    In reality, if the range of f (in my question) is [0, ∞), then ∫_{-n}^n f(x) dx = 0, and ∫_ℝ f(x) dx = 0, so f=0 a.e. is straightforward to see.

     – Jaborandi Kakapo Jun 02 '24 at 15:05
  • The answers provided a long time ago are too complicated. Just write $f=f_+-f_-$ and $F_{\pm }(x)=\int_a^x f_{\pm}(t)dt.$ Therefore $F_+(x)$ and $F_-(x)$ are equal and are the distribution functions of the same positive measure, implying $f_+=f_-$ almost everywhere. – Letac Gérard Jun 02 '24 at 17:10

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In order for $\int_a^b f(x)\; dx$ to be defined (as a Lebesgue integral), we need $f \in L^1(a,b)$. The Lebesgue differentiation theorem then says that for almost every $t$, $$ f(t) = \lim_{\varepsilon \to 0} \frac{\int_{t-\varepsilon}^{t+\varepsilon} f(x)\; dx}{2\varepsilon} = 0 $$

Robert Israel
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