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Suppose $f: \mathbb{R} \to \mathbb{R}$ is increasing and $g = f$ almost everywhere with respect to Lebesgue measure (a.e.). Suppose $g'$ exists a.e. Does it follow that $g' = f'$ a.e.?

This comes from a proof of the result that $f'$ exists almost everywhere when $f$ is increasing. Thus this result cannot be used.

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  • $\begingroup$ First, show that $f'$ exists a.e. by using $f=g$ a.e. and $g'$ exists a.e. Then, consider $g-f = 0$, so $(g-f)' = g'-f' = 0 \implies g'=f' \textrm{ a.e.}$ $\endgroup$ – Emily Apr 28 '15 at 19:13
  • $\begingroup$ @Arkamis Thanks! Unfortunately I'm still having trouble with the first part. For example, if $g = 1$ and $f = \mathbb{1}\{x \text{ is irrational}\}$, then $f'$ does not exist anywhere. I think the monotonicity assumption on $f$ gets us out of these types of situations, but am unsure how to use it. $\endgroup$ – David Apr 28 '15 at 19:27
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$f$ increasing implies $f'(x)$ exists a.e. So there will be a set $E\subset \mathbb {R}, m(\mathbb {R}\setminus E)=0,$ such that $f'(x),g'(x)$ exist, and $f(x)=g(x),$ for every $x\in E.$ Fix such an $x.$ Then there exists a sequence $x_n$ in $E$ with $x_n \to x.$ Thus

$$\frac{f(x_n)-f(x)}{x_n-x}= \frac{g(x_n)-g(x)}{x_n-x}$$

for all $n.$ This implies $f'(x) = g'(x)$ and we're done.

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  • $\begingroup$ Thanks for the response, but this comes out of a proof to show that $f'$ exists almost everywhere. I've amended the question to make this more clear. $\endgroup$ – David Apr 28 '15 at 20:34

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