Let $\mu$ be a positive measure in $\left(\mathbb{R},\mathcal{M}\right) $, $f$ and $g$ are two measurable functions equal to each other almost everywhere on $\left(\mathbb{R},\mathcal{M},\mu\right) $. Assume that $f$ and $g$ are continuous on $\mathbb{R}$. Does $f(x)=g(x) \forall x \in \mathbb{R}$?

I think the answer is no. Let $f=\chi_{A},g=0$ with $A=\left\{ 1\right\} $. We have $f=g$ a.e but $f(1)=1$, $g(1)=0$. I hope I don't have mistakes.

Thanks in advanced.

  • 2
    $\begingroup$ Your $f$ is not continuous. The answer is "it depends on $\mu$". $\endgroup$ – Daniel Fischer Dec 21 '13 at 14:45
  • $\begingroup$ Can you explain more precise? $\endgroup$ – chuyenvien94 Dec 21 '13 at 14:45
  • $\begingroup$ What does "almost everywhere" mean? Can you see the dependence on $\mu$? $\endgroup$ – Daniel Fischer Dec 21 '13 at 14:48
  • $\begingroup$ Chuy, from your questions here, I am thinking you are not ready to solve the problem quoted. If this is from a book you are reading on your own, then maybe you need to do a more basic book first. $\endgroup$ – GEdgar Dec 21 '13 at 15:53

The set $\{x\colon f(x)\ne g(x)\}$ is open, so if it is not empty then it contains a proper interval. If all of those have positive measure then the proposition holds.

  • $\begingroup$ Can you explain more precise? Why that set is open? $\endgroup$ – chuyenvien94 Dec 21 '13 at 14:52
  • 1
    $\begingroup$ This set is open because it is the preimage of the open set $\mathbb R\setminus\{0\}$ under the continuous function $f-g$. $\endgroup$ – Tomas Dec 21 '13 at 14:55
  • $\begingroup$ Ehm...of course, thank you @Tomas $\endgroup$ – chuyenvien94 Dec 21 '13 at 15:00
  • $\begingroup$ @Carsten Schultz: $\mu$ is a positive measure $\endgroup$ – chuyenvien94 Dec 21 '13 at 15:02
  • $\begingroup$ Can you prove the last statement: "If all of those have positive measure then the proposition holds". $\endgroup$ – chuyenvien94 Dec 21 '13 at 15:03

The answer is Not necessarily.

For example, a positive measure in $\mathbb R$ is the uniform measure $\mu$ in $[0,1]$, i.e., $$ \int_{\mathbb R} f\,d\mu = \int_0^1 f(x)\,dx. $$ With respect to this measure, $f=g$ a.e. iff $f(x)=g(x)$ a.e. in $[0,1]$, in the Lebesgue. measure sense. So if $f$ and $g$ are different for all $x\in\mathbb R\smallsetminus [0,1]$, then they are still equal a.e. with respect to $\mu$.

On the other hand, if $f=g$ a.e. with respect to the Lebesgue measure in $\mathbb R$, then they are equal everywhere.


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