# A question about lebesgue integral

pswongaa
if lebesgue integral of f^2 over an interval equal 0, must f=0 a.e on that interval?

R136a1
No. Try to find a counterexample (hint: the integral can be 0 since positive and negative parts cancel out).

Staff Emeritus
No. Try to find a counterexample (hint: the integral can be 0 since positive and negative parts cancel out).
What negative parts, R136a1? He's integrating f(x)2 over some interval.

if lebesgue integral of f^2 over an interval equal 0, must f=0 a.e on that interval?
Is f a function that maps reals to reals, or something else?

R136a1
Oh god. Never mind my reply.

Axiomer
For any measure space $(X,\mathcal{S},μ)$, and any measurable function $g:\rightarrow [-∞,∞]$, $$∫|g|dμ=0\implies g=0 a.e.$$

Specifically, since $f^2=|f^2|$, this gives $f^2=0$ a.e., and hence $f=0$ a.e.

Staff Emeritus
You are assuming f is a real function, Axiomer. If it's a complex function, then f2 is not the same as |f2|.

Axiomer
That's a good point. Since the op didn't specify otherwise, I assumed we were talking about functions to the extended real line.

Axiomer
For any measure space $(X,\mathcal{S},μ)$, and any measurable function $g:\rightarrow [-∞,∞]$, $$∫|g|dμ=0\implies g=0 a.e.$$

proof:
Define $A=\{x\in X: g(x)≠0\}$. For all naturals n, define $A_n=\{x\in X: |g(x)|>\frac{1}{n}\}$.

$\frac{1}{n}μ(A_n)=∫\frac{1}{n}x_{A_n}dμ≤∫|g|dμ=0$, so $μ(A_n)=0$ for all n.

Then $μ(A)=μ(\bigcup _{n=1}^∞A_n)≤\sum _{n=1}^∞μ(A_n)=0\implies μ(A)=0$, as desired.