Lebesgue Integral

**problem** · August 21st 2009, 06:34 AM

If f is measurable function such that $\int_B f =0$ for all measurable sets $B \subset A$ ,show that f =0 almost everywhere on $A$ .

**Opalg** · August 21st 2009, 12:10 PM

Originally Posted by problem

If f is measurable function such that $\int_B f =0$ for all measurable sets $B \subset A$ ,show that f =0 almost everywhere on $A$ .

For n=1,2,3,..., let $B_n = \{x\in A:f(x)\geqslant1/n\}$ . Then $B_n$ is a measurable subset of A, and so $\int_{B_n} \!\!f\,d\mu =0$ (where $\mu$ denotes the measure). But $\int_{B_n}\!\!f\,d\mu \geqslant\mu(B_n)/n$ . Therefore $\mu(B_n)=0$ . Since this holds for all n, it follows that $\textstyle\mu\Bigl(\bigcup_nB_n\Bigr)=0$ , which implies that f is nonpositive almost everywhere. Now do the same for the sets on which $f \leqslant-1/n$ , to conclude that f is nonnegative almost everywhere. Hence f=0 almost everywhere.

Thread: Lebesgue Integral

LinkBack

Thread Tools

Display