1. ## Lebesgue Integration

Let $(S,\Sigma,\mu)$ be a measure space. Prove that if $f$ is a Lebesgue integrable function, then for each $\varepsilon > 0$ there exists a $\delta > 0$ such that $\int_A \mid f \mid \mathrm{d} \mu$ for each $A \in \Sigma$ with $\mu(A) < \delta$.

2. Originally Posted by syme.gabriel
Let $(S,\Sigma,\mu)$ be a measure space. Prove that if $f$ is a Lebesgue integrable function, then for each $\varepsilon > 0$ there exists a $\delta > 0$ such that $\int_A \mid f \mid \mathrm{d} \mu \color{red}<\varepsilon$ for each $A \in \Sigma$ with $\mu(A) < \delta$.
Suppose on the contrary that for some $\varepsilon>0$ there exists, for each n=1,2,3,..., $E_n\in\Sigma$ such that $\mu(E_n)<2^{-n}$ but $\int_{E_n}|f|\,d\mu\geqslant\varepsilon$. Let $E = \limsup_nE_n = \bigcap_nF_n$, where $F_n = \bigcup_{k=n}^\infty E_k$. Then $\mu(E)\leqslant\sum_{i=n}^\infty\mu(E_i)<2^{-n+1}$ for all n, and hence $\mu(E)=0$.

On the other hand it follows from Dominated Convergence or some such theorem that $\int_E|f|\,d\mu = \lim_n\int_{F_n}|f|\,d\mu \geqslant\limsup_n\int_{E_n}|f|\,d\mu\geqslant\var epsilon$. That is a contradiction since clearly |f| cannot have a nonzero integral over a null set.

The idea of this proof comes from the fact that if we define a new measure $\nu$ on $\Sigma$ by $\nu(A) = \int_A|f|\,d\mu\ (A\in \Sigma)$, then $\nu\ll\mu$. There is then a standard theorem (Theorem 30B in Halmos's Measure Theory) saying that absolutely continuous measures have this "continuity" property. I just took Halmos's proof and re-wrote it for this situation.