Let $f$ be a non-negative measurable and integrable function on measure space $(X,M,\mu)$. Show that for every $\epsilon >0 $ there exists an $\delta >0 $ such that for every $E \in M$ with $\mu(E) \le \delta$ we have$\int\limits_Efd\mu \le \epsilon$
Have been trying to approach it via simple function approximation. However depending on different $E$ we get different simple functions. Hence a uniform bound on them is not possible. Any suggestions?
Best Answer
For every measurable $E$ and every $x\gt0$, $$ \int_Ef\mathrm d\mu=\int_{E\cap[f\lt x]}f\mathrm d\mu+\int_{E\cap[f\geqslant x]}f\mathrm d\mu\leqslant x\mu(E)+\int_{[f\geqslant x]}f\mathrm d\mu. $$ Choose $\varepsilon\gt0$. Since $f$ is integrable and $f\cdot\mathbf 1_{[f\geqslant x]}\leqslant f$, Lebesgue dominated convergence theorem shows that the rightmost integral in the RHS goes to zero when $x\to\infty$. Hence, there exists $x_\varepsilon$ such that this integral is $\leqslant\varepsilon/2$. Now, $\delta=\varepsilon/(2x_\varepsilon)$ guarantees that the LHS is $\leqslant\varepsilon$ for every $E$ such that $\mu(E)\leqslant\delta$.