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I am trying to show that if $f$ and $g$ are continuous functions on $[a, b]$ and if $f=g$ a.e. on $[a, b]$, then, in fact, $f=g$ on $[a, b]$. Also would a similar assertion be true if $[a, b]$ was replaced by a general measurable set $E$ ?

Some thoughts towards the proof

  1. Since $f$ and $g$ are continuous functions, so for all open sets $O$ and $P$ in $f$ and $g$'s ranges respectfully the sets $f^{-1}\left(O\right) $ and $g^{-1}\left(P\right) $ are open.
  2. Also since $f=g$ a.e. on $[a, b]$ I am guessing here implies their domains and ranges are equal almost everywhere(or except in the set with measure zero). $$m(f^{-1}\left(O\right) - g^{-1}\left(P\right)) = 0$$

I am not so sure if i can think of clear strategy to pursue here. Any help would be much appreciated.

Also i would be great full you could point out any other general assertions which if established would prove two functions are the same under any domain or range specification conditions.

Cheers.

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You can use continuity to show that if $f$ and $g$ take different values at $c$, then they take different values at every point of some (small) nhood of $c$. –  David Mitra Aug 1 '12 at 21:39
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3 Answers

up vote 10 down vote accepted

The set $\{x\mid f(x)\neq g(x)\}$ is open (it's $(f-g)^{—1}(\Bbb R\setminus\{0\})$), and of measure $0$. It's necessarily empty, otherwise it would contain an open interval.

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It's also worth thinking about what this argument says when you take complements: if a set has full (Lebesgue) measure (i.e. its complement has measure zero) then it is dense, and two continuous functions which agree on a dense set must agree everywhere. –  Nate Eldredge Aug 1 '12 at 22:05
    
@Davide Giraudo Just a quick question why did you exclude the set containing zero from R –  Hardy Aug 1 '12 at 22:11
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@Hardy Because $f(x)-g(x)=0$ means that $f$ and $g$ agree on $x$. We look at where they don't. –  Michael Greinecker Aug 1 '12 at 22:12
    
@NateEldredge Thank you very much for the other pointer. –  Hardy Aug 1 '12 at 22:12
    
@MichaelGreinecker but ofcourse, that make perfect sense. Thanks. –  Hardy Aug 1 '12 at 22:13
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If $f$ vanishes a.e. on $[a,b],$ there exists a measurable set $X$ s.t. $m(X)=0$ and $f=0$ on $[a,b]\setminus X.$
But $m(X)=0$ implies that $X$ has empty interior, (because any non empty open set has positive measure), or equivalently its complement is everywhere dense.

Now, from $f=0$ on $[a,b]\setminus X$ and $[a,b]\setminus X$ dense in $[a,b],$ we conclude that the continuous function $f$ vanishes everywhere on $[a,b].$

P.S. The same argument works for continuous functions on any Euclidean space.

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@Nate Eldredge I have corrected my wrong terminology. –  Giuseppe Tortorella Aug 1 '12 at 22:05
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A subset of an interval of full measure is dense in that interval. If two functions agree on a dense set, they are equal everywhere.

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