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Let $E$ be a subset of $\Bbb R$ with positive Lebesgue measure, $\lambda(E)>0$. Let $f$ be a function from $\Bbb R$ to $\Bbb R$ which is positive on $E$, that is $f(x)>0$ for all $x\in E$.

Is it possible that $$\int_E f\,d\lambda=0?$$ In other words, must $$\int_E f\,d\lambda$$ be strictly positive?

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    $\begingroup$ $\int_E f\,d\lambda=0 \iff f=0$ almost everywhere $\endgroup$ – iwriteonbananas Jan 28 '15 at 19:53
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    $\begingroup$ you can also approximate $f$ by simple functions. On a subset of $F\subset E$ you will find that $f_k\leq f$ and $f_k\geq \varepsilon >0$ $\endgroup$ – Quickbeam2k1 Jan 28 '15 at 20:05
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Since $f$ is strictly positive on $E$, we have $$ E = \bigcup_{n \geq 1} E_n, \quad \mbox{ where } E_n = \left\{x \in E: f(x) > \frac{1}{n}\right\}. $$ Since $\lambda(E) > 0$ there is some $n$ for which $\lambda(E_n)$ is positive (otherwise $E$ would be the countable union of measure $0$ sets, implying $\lambda(E)=0$). We then have $$ \int_E f \, d\lambda \geq \int_{E_n} f \, d\lambda > \int_{E_n} \frac1n \, d\lambda = \frac{\lambda(E_n)}{n} > 0, $$ as desired.

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Could I create a simple function K(x), where K(x) is 1 when x does not belong in E and 0 when x does belong in E?

If f(x)>0 this would imply that on the set E, f(x)>K(x) for all x belonging in E.

Further by the monotonicity nature of the lebesgue integral, the integral of f(x) is greater than the integral of K(x) which is 0.

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