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I'm reading Real Analysis by Royden (4th edition).

The problem I'm working on is in the title, were $f$ and $g$ are continuous functions on $[a,b]$.

What I know so far is that the set $\{x\,:\,f(x)\neq g(x)\}$ has measure zero. Also, I realized that $(f-g)^{-1}(R-\{0\})=\{x\,:\,f(x)\neq g(x)\}$. (R for the reals) What I'm thinking is to some how show that this set is empty, that way $f=g$ on all of $[a,b]$.

Thanks for any hints or feedback!

S.D.
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1 Answers1

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Note that $h = f - g$ is a continuous function which is zero almost everywhere. As $[a, b]$ is compact, $|h|$ must attain a maximum at some $x_0 \in [a, b]$. If $|h(x_0)| > 0$, there exists a neighborhood $(x - \delta, x + \delta)$ on which $|h| > \frac{1}{2}|h(x_0)| > 0$, by continuity; but this interval has positive Lebesgue measure $2\delta$. This contradiction shows that $|h|$ cannot have a positive maximum, and so must be identically zero.