6

This is exercise 4.18 from "A User-Friendly Introduction to Lebesgue Measure and Integration" by Gail S. Nelson:

Let $(X,\mathcal{B},\mu)$ be a complete measure space. Let $f$ be integrable with respect to the measure $\mu$. For $A\in\mathcal{B}$ we define $\int_{A} f d\mu=\int f\cdot\mathcal{X}_{A} d\mu$. Prove: Given $\epsilon>0$ there is $\delta>0$ such that if $A\in\mathcal{B}$ and $\mu(A)<\delta$, then $\Big|\int_{A} f d\mu\Big|<\epsilon$.

Note: here, $\mathcal{X}_{A}(x)$ is characteristic function of set $A$, that is equal $1$ if $x\in A$ and equal $0$ if $x\notin A$.

At the moment, I really don't have an idea on how to start the proof. Some additional definitions: A function is integrable with respect to the measure $\mu$ if its positive and negative parts are integrable, and non-negative function $f$ is integrable if $\sup\Big\{\int \Phi d\mu\ \Big|\ \textrm{$\Phi$ is a simple function with $0\le\Phi\le f$}\Big\}$ is finite (and integral $\int f d\mu$ is then equal to this supremum). A simple function $\Phi(x)$ is a function defined as $\Phi(x)=\sum_{k=1}^{n}a_{k}\cdot\mathcal{X}_{E_{k}}(x)$, where $a_{k}$ are real constants and $E_{k}$ are pairwise disjoint measurable sets.

HallaSurvivor
  • 38,115
  • 4
  • 46
  • 87
Crni
  • 647

1 Answers1

13

Assume that $f \ge 0$. If $A$ is measurable and $\Phi$ is a simple function with $0 \le \Phi \le f$ then $$\int_A(f - \Phi) \, d\mu \le \int (f-\Phi) \, d\mu$$ so that $$\int_A f \, d\mu \le \int_A \Phi \, d\mu + \int f \, d\mu - \int \Phi \, d\mu.$$

Let $\epsilon > 0$ be given. You may choose $\Phi$ so that $$\int f \, d\mu - \int \Phi \, d\mu < \frac{\epsilon}2.$$ Let $M = \sup \Phi$, and let $\delta = \frac{\epsilon}{2M}$. Then $\mu(A) < \delta$ implies $$\int_A \Phi \, d\mu \le M \mu(A) < \frac{\epsilon}2$$ so that $\mu(A) < \delta$ implies $$\int_A f \,d\mu < \epsilon.$$

Umberto P.
  • 52,165