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!

 

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