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Here is Prob. 1, Chap. 3, in the book Real Analysis by H. L. Royden and P. M. Fitzpatrick, 4th edition:

Suppose $f$ and $g$ are continuous functions on $[a, b]$. Show that if $f = g$ a.e. on $[a, b]$, then, in fact, $f = g$ on $[a, b]$. Is a similar assertion true if $[a, b]$ is replaced by a general measurable set?

My Attempt:

Let $E_0$ be a subset of $[a, b]$ such that $m^* \left( E_0 \right) = 0$ and $f(x) = g(x)$ for all $x \in [a, b] \setminus E_0$.

What next? How to proceed from here?

Saaqib Mahmood
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    https://math.stackexchange.com/q/543648/42969, https://math.stackexchange.com/q/1944857/42969 – Martin R Jun 05 '21 at 20:52
  • The set where two continuous functions agree is a closed set. – GEdgar Jun 05 '21 at 20:52
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    You still haven't made any comment to the detailed answer I have done here to one of your recent questions (I was the unique answerer). Consider that people who spend time to solve your questions desire to have some echo, at least some days later... – Jean Marie Jun 05 '21 at 21:09
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    @JeanMarie I'm really sorry but earlier on I just couldn't examine your post carefully enough so as to be able to give any feedback. – Saaqib Mahmood Jun 06 '21 at 14:25

2 Answers2

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The set $[0,1]\cup\{2\}$ is measurable. Let $f(x) = x$ for $x$ in this set, and let $g(x)= x$ for $x\in[0,1]$ and $g(2)=3.$ Then $f$ and $g$ agree almost everywhere on their domain, and both are continuous on that domain, but they don't agree everywhere.

Michael Hardy
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Let $E=\{x\in [a,b]:f(x)\ne g(x)\}$. Since $E$ has measure $0$, for each $x\in [a,b]$, there exists a sequence $\{x_n\}$ in $[a,b]\setminus E$ such that $x_n\to x$. (Why?)

Since $f$ and $g$ are continuous, $$0 = \lim_{n\to\infty}[f(x_n)-g(x_n)] = f(x)-g(x).$$

Alex Ortiz
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  • can you please also answer the (Why?) in your post? – Saaqib Mahmood Jun 06 '21 at 14:30
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    In order to find such a sequence, we have to show that, for every $n\in \mathbb{N}$, the set $(x-1/n, x+1/n) \cap [a,b]$ contains a point $x_n$ which doesn't lie in $E$. What would happen if that weren't the case for one $n\in \mathbb{N}$? – Mandelbrot Jun 06 '21 at 17:10