# $f=g$ almost everywhere $\int fdm= \int gdm$ [closed]

Let $f$ and $g$ be positive measurable functions. Show that if $f=g$ almost everywhere, then $\int f\,dm= \int g\,dm$.

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## closed as off-topic by Shaun, Davide Giraudo, drhab, Jack, Guy FsoneNov 26 '17 at 14:33

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• Where are you having difficulties? What have you tried? – RideTheWavelet Nov 26 '17 at 14:18
• You have that $h=f-g=0$ almost everywhere. What about $\int h\,dm$? – egreg Nov 26 '17 at 14:26

Let $(\Omega,\mu)$ be a measurable space and $A=\{x\in\Omega\colon f(x)\ne g(x)\}$. Then $\mu(A)=0$ by our assumption. We have $$\int\limits_{\Omega}f\,\text{d}m=\int\limits_{\Omega\setminus A} f\,\text{d}m+\int\limits_{A}f\,\text{d}m=\int\limits_{\Omega\setminus A} f\,\text{d}m$$ since the integral over a set of mearure zero vanishes. What about $f$ and $g$ on $\Omega\setminus A$?
By the way, the assumptiom that $f,g$ are positive is redundant here. Measurability is enough.