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Let f be a non-negative Lebesgue measurable function with $\int_\mu\ f \,d\lambda< \infty$. Show that for each $\epsilon>0$, there exists a $\delta>0$ such that $\lambda(E)<\delta$ implies $\int_E\ f \,d\lambda< \epsilon$.

--I think I need to use the simple function $\phi=f\chi_E$ such that $\phi<f$ but I do not see how to tie that into the standard epsilon-delta proof with a finite l Lebesgue integral.

--I also know that $\lambda(A_c)\leqslant(1/c)\int_E\ f \,d\lambda$ from the previous problem so I am assuming I will need to use that.

Rashelle DeBolt
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1 Answers1

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Set $f_n(x)=\begin{cases} n &f(x)\ge n \ \\ f(x) & 0\leq f(x)<n \\ \end{cases}$

Then, for any measurable $E,\ \int_E f=\int_E (f-f_n)+\int_E f_n\le \int_E (f-f_n)+n\cdot \lambda(E)$.

By the Monotone Convergence Theorem $\int_E (f-f_n)\to 0$ as $n\to \infty,$ so if we choose $N$ large enough so that $\int_E (f-f_N)<\epsilon/2$ and take $\delta <\epsilon/2N$ then if $\lambda (E)<\delta,\ \int_E f<\epsilon.$

Matematleta
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