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The title says it all.

I've already shown, for an earlier part of this problem, that for any $E$ with $\lambda(E)>0$, we have $\int_E fd\lambda >0$. I did that by reductio, showing that assuming the contrary allows us to prove $f=0$ a.e.; I'm not sure how to use that here.

So far I've tried another reductio, assuming that there exists a sequence of measurable functions $f_n$ such that $f_n=f\cdot\mathbf 1_{E_n}$, where $E_n$ is such that $\int_{E_n}fd\lambda \leq \frac1n$ and $\lambda(E_n)\geq\epsilon$. This gives us $\lim_{n\to\infty}\int_{[0,1]}f_nd\lambda=0$, but I can't see where to go from here. Any pointers?

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Since $f$ is non-negative, we have for each $n$, \begin{align}\int_Ef(x)\mathrm d\lambda(x)&\geqslant \frac 1n\lambda\left(E\cap \{f\gt 1/n\}\right)\\ &=\frac 1n\left(\lambda\left(E\right)-\lambda\left(E\cap \{f\leqslant 1/n\}\right)\right)\\ &\geqslant \frac 1n\left(\lambda\left(E\right)-\lambda\left( \{f\leqslant 1/n\}\right)\right). \end{align} Now, if $\varepsilon$ is fixed, choose $n$ such that $\lambda\left( \{f\leqslant 1/n\}\right)\lt\varepsilon/2$ in order to get $$\int_Ef(x)\mathrm d\lambda(x)\geqslant \frac{\varepsilon}{2n}$$ for each $E$ such that $\lambda(E)\geqslant\varepsilon$.

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