0
$\begingroup$

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

Could you explain this question?

$\endgroup$

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shaun, Davide Giraudo, drhab, Jack, Guy Fsone
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Where are you having difficulties? What have you tried? $\endgroup$ – RideTheWavelet Nov 26 '17 at 14:18
  • 1
    $\begingroup$ You have that $h=f-g=0$ almost everywhere. What about $\int h\,dm$? $\endgroup$ – egreg Nov 26 '17 at 14:26
2
$\begingroup$

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.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.