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Let $(X, \mathcal{B}, \mu)$ be a measure space and $f$ a measurable function on $X$ and suppose that $\forall E \in \mathcal{B}$ we have that $\int_E f = 0$. Then I want to show that $f = 0$ almost everywhere (a.e.).

  1. Suppose for sake of contradiction that $f \ne 0$ a.e.
  2. Then $\not\exists E \in \mathcal{B}$ s.t. $\mu(E) = 0$ and $f(x) = 0$, $\forall x \in X - E$

  3. Then $\{x : f(x) \ne 0\} = A$ is not measure zero so that either $\mu(A) > 0$ or $A \notin \mathcal{B}$.

  4. Now if $\mu(A) > 0$ then it is easy to see that $\int_A f \ne 0$ so that we have a contradiction of our original hypothesis that $\forall E \in \mathcal{B}, \int_E f = 0$.

  5. But if on the other hand $A \notin \mathcal{B}$, I cannot no longer appeal to $\int_A f \ne 0$ since $\int_A f$ is non-sense. So I'm having trouble with this part of the argument.

George
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2 Answers2

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Arguing by contradiction definitely works, here's the idea.

Let $\mu(\{ x : f(x) \neq 0 \}) > 0$.

Then we have $\{x : f(x) \neq 0\} = \{x : f(x) > 0\} \cup \{x : f(x) < 0\}$, so we must have one of these two sets have positive measure.

Let's say its the first one (the argument for the second is analogous).

Then $\{x : f(x) > 0\} = \bigcup \{ x : f(x) \geq \frac{1}{n}\} = \bigcup E_n$ so again one of these must have positive measure.

So say $E_k$ has positive measure, then $f$ dominates $\frac{1}{k}$ on $E_k$ so $$\int_{E_k} f \geq \int_{E_k} \frac{1}{k} = \mu(E_k)\frac{1}{k} > 0.$$ So we have a contradiction.

EDIT: also, to comment on your proof: as you see here we don't need to deal with whether or not $A$ is measurable, it definitely is. And also, for step $4$ of your proof $\mu(A) > 0$ does not imply that $\int_A f \neq 0$ for example if $f = 1_{(0,1)} - 1_{(-1,0)}$

Deven Ware
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  • Is it possible to prove this without using contradiction? I've been thinking about how to do it this way, but I can't come up with a way to do it. Thanks! – user110320 Mar 07 '18 at 18:45
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    Pardon me if I am mistaken, but don't we actually need to assume that $f$ is non-negative for this result, because it strikes me that the argument for the second case does not go through analogously. – Rio Alvarado Dec 03 '19 at 07:03
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The set $\{x:f(x)\neq 0\}$ is the same as $f^{-1}\big((-\infty,0)\big)\cup f^{-1}\big((0,\infty)\big)$ and therefore measurable. So the issue in 5. never occurs.

And I think you should be more explicit in step 4.

Michael Greinecker
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