## Interesting Analysis Question (Measure Theory)

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Let $f\geq 0$ be a measurable function with $\int f d\mu<\infty$. Show that for any $\epsilon>0$, there exists a $\delta(\epsilon)>0$ such that for any measurable set $E\in\mathcal{A}$ with $\mu(E)<\delta(\epsilon)$, we have $\int_E f d\mu<\epsilon$.

Proof: For $M>0$, we define $f_M(x)=\min (f(x),M)\leq M$, for all $x\in \Omega$.

Then $f=f_M+(f-f_M)$.

Let $\delta=\epsilon/2M$. Then for any $E\in\mathcal{A}$ with $\mu(E)<\delta$,

\begin{aligned}\int_E f_M d\mu &\leq \int_E M d\mu\\ &=\mu (E)M\\ &<\delta M\\ &=(\epsilon/2M)M\\ &=\epsilon/2 \end{aligned}

Note that $f_M\uparrow f$. By Monotone Convergence Theorem,

$\int f d\mu=\lim_{M\to\infty}\int f_M d\mu$.

Therefore $\lim_{M\to\infty}\int f-f_M d\mu=0$.

We can choose $M$ sufficiently large such that

$\int_E f-f_M d\mu \leq \int f-f_M d\mu <\epsilon/2$.

Then

\begin{aligned} \int_E f d\mu&=\int_E f_M d\mu+\int_E f-f_M d\mu\\ &<\epsilon/2+\epsilon/2\\ &=\epsilon \end{aligned}

We are done!