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Background Information:

This is a Corollary to Theorem 3.5, found here.

If $\mu$ is a measure and $f$ is an extended $\mu$-integrable function, the signed measure $\nu$ defined by $\nu(E) = \int_{E}f d\mu$ is clearly absolutely continuous w.r.t. $\mu$; it is finite iff $f\in L^1(\mu)$.

Corollary 3.6 - If $f\in L^1(\mu)$, for every $\epsilon > 0$ there exists a $\delta > 0$ such that $\left|\int_{E}f d\mu\right| < \epsilon$ whenever $\mu(E) < \delta$.

Attempted proof - First let us prove that $\nu(E) = \int_E f d\mu$ is a measure. Clealy, $\nu(\emptyset) = \int_{\emptyset} f d\mu = \int f\chi_{\emptyset}d\mu = \int f\cdot 0 d\mu = 0$. Now, let $\{E_j\}_{1}^{\infty}$ be a sequence of disjoint sets in $M$. Then, $$\nu\left(\bigcup_{1}^{\infty}E_j\right) = \int_{\bigcup_{1}^{\infty}E_j}f d\mu = \int\left(\sum_{1}^{\infty}\chi_{E_j}\right)f d\mu = \sum_{1}^{\infty}\int f\chi_{E_j} d\mu = \sum_{1}^{\infty}\nu(E_j)$$ Now if there exists a set $E\in M$ such that $\mu(E) = 0$ then $\int_E d\mu = 0$ and so $\int_E f d\mu = 0$ thus $\nu(E)$ is absolutely continuous. Then we can use theorem 3.5 to conculde that $\left|\int_E f d\mu\right| < \epsilon$ whenever $\mu(E) < \delta$.

I am not exactly sure how to do the complex part when $f$ is a complex-valued function. Any suggestions is greatly appreciated.

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  • $\begingroup$ A complex-valued function is a linear combination of two real-valued functions. Also $|x+iy|\le|x|+|y|$. $\endgroup$ – David C. Ullrich Jul 12 '16 at 22:48
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@Wolfi , your proof is correct for the case of $f$ real-valued function. I have copied it here, add minor details for clarification and complete it.

Corollary 3.6 - If $f\in L^1(\mu)$, for every $\epsilon > 0$ there exists a $\delta > 0$ such that $\left|\int_{E}f d\mu\right| < \epsilon$ whenever $\mu(E) < \delta$.

Proof - Let $f\in L^1(\mu)$. First, let us prove that $\nu(E) = \int_E f d\mu$ is a measure. Clealy, $\nu(\emptyset) = \int_{\emptyset} f d\mu = \int f\chi_{\emptyset}d\mu = \int f\cdot 0 d\mu = 0$. Now, let $\{E_j\}_{1}^{\infty}$ be a sequence of disjoint sets in $M$. Then, $$\nu\left(\bigcup_{1}^{\infty}E_j\right) = \int_{\bigcup_{1}^{\infty}E_j}f d\mu = \int\chi_{\bigcup_{1}^{\infty}E_j}f d\mu =\int\left(\sum_{1}^{\infty}\chi_{E_j}\right)f d\mu = \sum_{1}^{\infty}\int f\chi_{E_j} d\mu = \sum_{1}^{\infty}\nu(E_j)$$ Now for any set $E\in M$, if $\mu(E) = 0$ then $\int_E f d\mu = 0$ thus $\nu(E)$ is absolutely continuous with respect to $\mu$.

Suppose that $f$ is real-valued, then we can use theorem 3.5 to conclude that for all $\epsilon >0$, there is $\delta >0$ such that $\left|\int_E f d\mu\right| < \epsilon$ whenever $\mu(E) < \delta$.

Now, suppose $f$ is a complex-valued function, then we have that $\textrm{Re}f \in\in L^1(\mu)$ and $\textrm{Im}f \in\in L^1(\mu)$. So applying to $\textrm{Re}f$ and $\textrm{Im}f$ what we have just prove, we have that:

  1. for all $\epsilon >0$, there is $\delta_1 >0$ such that $\left|\int_E \textrm{Re}f d\mu\right| < \epsilon/2$ whenever $\mu(E) < \delta_1$.

  2. for all $\epsilon >0$, there is $\delta_2 >0$ such that $\left|\int_E \textrm{Im}f d\mu\right| < \epsilon/2$ whenever $\mu(E) < \delta_2$.

So taking $\delta = \textrm{min}\{\delta_1,\delta_2\}$, we have \begin{align*} \left|\int_E f d\mu\right|&=\left|\int_E (\textrm{Re}f + i \textrm{Im}f) d\mu\right|=\\&=\left|\int_E \textrm{Re}f + i\int_E \textrm{Im}f d\mu\right|\leq \\& \leq \left|\int_E \textrm{Re}f d\mu\right|+\left|\int_E \textrm{Im}f d\mu\right|<\\ & <\epsilon/2 +\epsilon/2 =\epsilon \end{align*} So we have proved that for all $\epsilon >0$, there is $\delta >0$ such that $\left|\int_E f d\mu\right| < \epsilon$ whenever $\mu(E) < \delta$.

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