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Suppose $f$ and $g$ are continuous functions on $[a,b]$. Show that if $f=g$ almost everywhere 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 $E$?

I have known that the set $A=\{x \mid f(x) \neq g(x)\}$ has measure zero and we want to show that A is empty. Now let's assume $A$ is not empty. I am stuck in getting the contradiction.

Thanks for your hints and answers.

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    $\begingroup$ Funny: this old thread bumped earlier today. $\endgroup$
    – Julien
    Oct 29, 2013 at 2:46
  • $\begingroup$ TeX tip: Use \mid (instead of just |) for the set builder to get correct spacing around it. $\endgroup$
    – kahen
    Jan 16, 2015 at 15:50

5 Answers 5

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You can do it even more simply:

$f,g$ continuous on $[a,b]$ implies $(f-g)$ continuous. Thus:

$$(f-g)^{-1}(0, + \infty)\cup(0, -\infty) = A$$

is open, thus $A=\emptyset$ as that is the only open set of measure zero. QED. No epsilonics required.

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Can I prove like this:

Assume that $A\neq \emptyset$. There exists $x_0\in[a,b]$ such that $f(x)-g(x)\neq 0$, WLOG, say $f(x)-g(x)=a>0$. Since $f,g$ are continuous, $f-g$ is also continuous. By the property of continuous function, for any $\epsilon$ there exists $\delta$ such that $|x-x_0|<\delta$ then $|(f-g)(x)- a|=|(f-g)(x) -(f-g)(x_0)|<\epsilon.$ By choosing $\epsilon = \frac{a}{2}$, $|(f-g)(x)-a|<\frac{a}{2}\to 0<\frac{a}{2}<(f-g)(x)< \frac{3a}{2}$. That means $f\neq g$ for all $x\in \{x:|x-x_0|<\delta\}\subset A$. Since $m \{x:|x-x_0|<\delta\}=2\delta>0$, we have a contradiction.

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A nonempty open subset of $[a,b]$ always has positive measure. Does that help?

More Hints: The inverse image of an open set under a continuous function is _ _ _ _? A nonempty open set always contains an open interval.

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  • $\begingroup$ Thanks though. Which set is the open subset of [a,b] and why is your statement true?@Amitesh $\endgroup$
    – Yang
    Oct 29, 2013 at 3:05
  • $\begingroup$ I've added further hints, @user98517. I hope they help but I think it's best if you think about this problem by yourself a little more. $\endgroup$ Oct 29, 2013 at 3:08
  • $\begingroup$ The inverse image of an open set under a continuous function is an open set. I appreciate your answers. I will think about it. $\endgroup$
    – Yang
    Oct 29, 2013 at 3:10
  • $\begingroup$ OK @user98517. Is it possible to find a nonempty open subset of, say, $\mathbb{R}$ with measure zero? (It may be helpful to recall some familiar open sets to answer this question.) $\endgroup$ Oct 29, 2013 at 3:11
  • $\begingroup$ I think it is possible@Amitesh $\endgroup$
    – Yang
    Oct 29, 2013 at 3:19
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Hint: Think about topological properties of $A$ and you'll get your contradiction.

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If $f = g$ almost everywhere on $[a,b]$ then $|f - g| = 0$ almost everywhere on $[a,b].$ Let $\mu$ denote the Lebesgue measure on $[a,b].$ Then $$\int_{a}^{b} |f-g|\ dx = \int_{[a,b]} |f-g|\ d\mu = 0.$$ Since $f,g \in C[a,b]$ it follows that $|f - g| \in C[a,b]$ and moreover $|f-g| \geq 0$ on $[a,b].$ So $$\int_{a}^{b} |f-g|\ dx = 0 \iff |f-g| \equiv 0 \iff f\equiv g.$$

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