# Integral vanishes on all intervals implies the function is a.e. zero

I am having trouble with the following problem:

$f:\mathbb{R}\to \mathbb{R}$ is a measurable function such that for all $a$: $$\int_{[0,a]}f\,dm=0.$$ Prove that $f=0$ for $m$ almost every $x$ (here $m$ is the Lebesgue measure).

I have no problem proving this for $f$ non-negative, or under the assumption that $f$ is integrable. But the question only assumes that $f$ is measurable and no more.

My idea was the usual thing; we look at the set of points where $f$ is positive and negative and assume one of these has measure greater than zero. Then I wanted to estimate one of these by an open set, look at the integral on the open set and show that it had to be greater than zero, a contradiction. But a key part of this attack is the assumption of the absolute continuity of the integral, which only holds in the case where $f$ is integrable.

Alternatively, if it were integrable one could simply estimate $f$ by a continuous function, where the result is quite obvious.

Ultimately we are going to show that $f$ is integrable, but it is not clear to me how to show this before showing it is zero a.e. So there must be a simpler way. Does anyone have suggestions?

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Do you mean to integrate over $[0,a]$? Otherwise one can take $f(x)=1$ for $x<0$ and $f(x)=0$ for $x\geq0$. Then the hypothesis is true while the conclusion fails. – Clayton Jan 9 '13 at 19:55
It says for all $a$, which I take to mean the interval $[a,0]$ when $a<0$ and $[0,a]$ when $a>0$. – rondo9 Jan 9 '13 at 19:57
I see. I took it to mean $\varnothing$ for $a<0$. I'll consider what you've said, though. – Clayton Jan 9 '13 at 19:58
Is the function assumed locally integrable? – Davide Giraudo Jan 9 '13 at 20:34
The assumptions are only that it is measurable and that its integral vanishes on intervals. – rondo9 Jan 9 '13 at 20:42

The function $f$ must be integrable (one of $\int f^+$ or $\int f^-$ is finite) in order for the symbol $\int f$ to be defined. So, I'll assume this is the case. In fact, then, since $\int_0^a f$ exists and is finite for any $a$, it follows that $\int_c^d |f|<\infty$ for any numbers $c$, $d$.

We show $f$ is almost everywhere $0$ on any interval $[c,d]$; this will imply the desired result.

Suppose $f>0$ on the set of positive measure $E\subset[c,d]$. Choose a closed subset $F$ of $E$ with positive measure. We then have $\int_F f>0$. Now let $U=[c,d]\setminus F$. As $U$ is open, we may write $U$ as a disjoint union of open intervals: $U=\bigcup_{k=1}^\infty (a_k,b_k)$.

Now, since $\int_c^d |f|<\infty$ $$0=\int_{[c,d]}f=\sum_{k=1}^\infty\int_{a_k}^{b_k}f+\int_F f.$$ Since $\int_F f>0$, it follows that $\sum\limits_{k=1}^\infty\int_{a_k}^{b_k}f$ is negative. But then $\int_{a_n}^{b_n} f$ must be negative for some $n$. However, this proves untenable upon observing that $$\int_{a_n}^{b_n} f =\int_0^{b_n} f - \int_0^{a_n} f =0.$$

Similarly, one can show $f$ cannot be negative on a set of positive measure.

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Thanks! I was totally overlooking that it IS locally integrable for the question to even mean anything. – rondo9 Jan 10 '13 at 0:00

Dynkin's $\pi-\lambda$ System theorem, as a strategic approach, works perfectly to conclude that the problem is true for all measurable sets !

Let $\mathcal{A}=$*"set of intervarls"* which is a $\pi$*-system* and $\mathcal{L}=\{A\in\mathcal{F};\;\int_A f=0\}$ which is a $\lambda$*-system*. Hence, $$\mathcal{B}=\sigma(\mathcal{A})\subseteq\mathcal{L}$$ It means for each Borel-set $B$ : $\displaystyle \int_B f = 0$

Now use this fact that any measurable set differs from a Borel-measurable set by a Zero-set.

To finish prove see : Showing that $f = 0$ a.e. if for any measurable set $E$, $\int_E f = 0$

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