# Thread: Lebesgue Integration

1. ## Lebesgue Integration

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

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

On the other hand it follows from Dominated Convergence or some such theorem that $\displaystyle \int_E|f|\,d\mu = \lim_n\int_{F_n}|f|\,d\mu \geqslant\limsup_n\int_{E_n}|f|\,d\mu\geqslant\var epsilon$. 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 $\displaystyle \nu$ on $\displaystyle \Sigma$ by $\displaystyle \nu(A) = \int_A|f|\,d\mu\ (A\in \Sigma)$, then $\displaystyle \nu\ll\mu$. 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.