Given two arbitary Lebesgue integrable functions we have,

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


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

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"

  • 5
    $\begingroup$ This "property of the integral" isn't trivial for this problem, I think. $\endgroup$ – Vim Sep 7 '17 at 17:20
  • $\begingroup$ Such exercise would depend on what theorems/propositions are allowed to use. $\endgroup$ – Jack Sep 7 '17 at 17:34
  • $\begingroup$ 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" $\endgroup$ – user415535 Sep 7 '17 at 17:36

You need this lemma to complete your proof:

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 $$\int h\ge\int_{B_k}h\ge\int_{B_k}\frac1k>0$$


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