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Hey guys I noticed this question was posted on this site earlier so I tried to work it out myself.

Let f:R→R be a bounded Lebesgue measurable function such that $\int_{[a,b]} f=0$ for all real a,b. Show that $\int_{E} f=0$ for each subset E of R of finite Lebesgue measure.

I was able to prove this is true for any non negative measurable function and did not even need that f be bounded. Is my proof correct and how would you extend it for any f not necessarily non negative? Thanks!

proof)By contradiction, assume there is an E measurable set of finite measure such that integral of f over E is strictly bigger than 0. Now E measurable implies that given any fixed epsilon there exists an open set O containing E s.t. m(O)$ \leq $m(E) + $\epsilon$ Hence O has finite measure as well. Now O contains E which implies $\int_{O} f \geq \int_{E} f$ . Now O open means we can write O as disjoint union of open intervals, say $O= \bigcup\limits_{i=1}^{\infty} (a_i,b_i)$ . This implies $\int_{\bigcup\limits_{i=1}^{\infty} (a_i,b_i)} f$>0 which implies $\int_R f \sum_{i=1}^\infty 1_{[a_i,b_i]} (x) dx = \sum_{i=1}^\infty \int_{[a_i,b_i]} f(x) dx > 0$ (since integrand is non negative, we can exchange infinite sum and integral)

But this implies since f is non negative that there is at least one i such that $\int_{[a_i,b_i]} f >0$ which is a contradiction.

Is my proof correct and is there a way to extend it to the general f case? I am thinking we might need to use dominated convergence? Thanks

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    "I noticed this question was posted on this site earlier so I tried to work it out myself." Link, please? – 2017-02-03
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    f is non negative and so integrating over a bigger set gives a bigger integral? – 2017-02-03
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    @Socchi I am sorry I just realized this was the case. Thank you for the point out. – 2017-02-03
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    no problem! and Jonas it was posted 7 or 8 days ago, I just closed the window and can not seem to find it anymore. In that post no solution was given. – 2017-02-03
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    @Socchi But then this solution is fine, isn't it? – 2017-02-03
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    That's what I wanted to verify, and also the question is to prove it for general f.. not assuming f is non negative. Only that it is bounded. – 2017-02-03

2 Answers 2

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Let $F(x) = \int_{[0,x]} f$, then $F = 0$ and the Lebesgue differentiation theorem gives $F'(x) = f(x) $ ae. $x$. Hence $f=0$ for ae. $x$. It follows that $\int_E f = 0$.

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    Oh wow that's neat, but just to make sure, where is the boundedness condition of f used? Is that used to justify that f is locally integrable, hence integrable on [0,x] to be able to use Lebesgue's differentiation theorem? Thanks. – 2017-02-03
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    Yes. ${}{}{}{}{}$ – 2017-02-03
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    So basically we showed that if f integrates to 0 on any finite interval, then f is identically 0 almost eveywhere? But then integral over any measurable E would be 0 too would it not? We do not need E to be finite measurable? – 2017-02-03
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    It is zero for any measurable $E$. – 2017-02-03
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Without the LDT: Suppose $|f|\le M$ on $\mathbb R.$ Let $E\subset \mathbb R$ have finite measure. Let $\epsilon>0.$ Then there is an open set $U\subset \mathbb R$ such that $E\subset U$ and $m(U\setminus E) < \epsilon.$ Now $U$ is the pairwise disjoint union of countably many open intervals $(a_n,b_n),$ and $\int_{a_n}^{b_n} f = 0$ for all $n.$ This implies $\int_U f=0.$ So

$$0 = \int_U f = \int_E f + \int_{U\setminus E}f. $$

The last integral in absolute value is $\le M\cdot\mu(U\setminus E) < M\epsilon.$ Therefore $|\int_E f| < M\epsilon.$ Since $\epsilon$ is arbitrary, $|\int_E f| =0$ as desired.

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    Hi, how did you get that integral over U of f=0? That is essentially saying that you can switch integral and infinite sum of indicator functions isn't it? – 2017-02-03
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    It's a DCT argument: Let $U_n = \cup_{k=1}^n (a_k,b_k).$ We certainly have $\int_{U_n}f =0.$ See if you can show, using DCT, that $\int_{U_n}f \to \int_{U}f.$ – 2017-02-03