TITLE: If $\int_{[a,b]} f=0$ , then $\int_E f=0 $ for any measurable $E \subseteq \mathbb{ R}$
QUESTION [0 upvotes]: 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

REPLY [0 votes]: 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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