# $\int f = \int g$ and $f\le g$ $\implies f=g$ a.e.

Given two arbitary Lebesgue integrable functions we have,

$$\int f dx= \int g dx$$ and $$f\le g \implies f=g$$ a.e.

Proof

$$\int g dx=\int f dx \implies \int g dx-\int f dx=0 \implies \int( g-f) dx=0$$

since $$g-f\ge 0$$ we know $$f=g$$ by property of the integral.

This proof looks soild to me but I cant really convice myself considering the "area under the curve"

• This "property of the integral" isn't trivial for this problem, I think. – Vim Sep 7 '17 at 17:20
• Such exercise would depend on what theorems/propositions are allowed to use. – Jack Sep 7 '17 at 17:34
• Well I used this "fact" as a part of some other proof and when I tought about it I got confused regarding the "area under the curve" – user415535 Sep 7 '17 at 17:36

If $h\ge 0$ is Lebesgue measurable, then $$\int h=0\implies h=0\,\text{a.e.}$$
Proof: suppose there exists $A$ with $m(A)>0$ and $h|_A>0$, then $$A=\cup_n (A\cap h^{-1}((\frac1n,\infty]))$$ Since $$m(A)=\lim_{n\to\infty}m(A\cap h^{-1}((\frac1n,\infty]))>0$$ There exists some $k$ such that $$m(A\cap h^{-1}((\frac1k,\infty]))>0$$ Let $B_k:=A\cap h^{-1}((\frac1k,\infty])$, then $$\int h\ge\int_{B_k}h\ge\int_{B_k}\frac1k>0$$