3

"Suppose $f$ and $g$ are continuous on the measurable set $E$. If $f = g$ a.e. on $E$, then $f=g$ on $E$."

I have shown that this statement is true when $E = [a,b]$ and I have shown that the statement is true for $E$ when $m(E)\neq 0$. However, my proof for the latter case never explicitly used the fact that $m(E)\neq 0$.

I have been told that the statement does not hold when $m(E)=0$. However, I am unconvinced (seems to me it should be vacuously true) and unable to prove it.

Could somebody please share with me a quick, not too complicated proof for why the statement is not true when $m(E) = 0$?

Thank you.

4 Answers4

3

*

I have shown that this statement is true when $E=[a,b]$ and I have shown that the statement is true for $E$ when $m(E) \not=0$.

*

Well done on the former. As for the latter ... not so good.

The principle at work here has little to do with measure theory. It is simply this: if $f$ and $g$ are continuous functions that agree on a dense subset of $E$ then they agree on all of $E$.

So if $N\subset [a,b]$ is a set of measure zero then you can use the fact that $[a,b]\setminus N$ is a dense subset of $[a,b]$.

If, however, $N\subset E$ is a set of measure zero it does not follow that $E\setminus N$ is a dense subset of $E$.

For example take $E=[0,1] \cup \{2\}$ and $N = \{2\}$. $N$ is a measure zero set but you simply cannot conclude that two continuous functions that agree on $[0,1]$ would have to agree on $E$.

B. S. Thomson
  • 7,463
  • Thompson, so it only worked for $[a, b]$ because finitely many points in $[a,b]$ are dense in $[a,b]$ (I suppose because of the density of the real numbers)? –  Oct 28 '15 at 15:50
  • Say it this way: if you remove a set of measure zero from an interval $[a,b]$ you leave behind a set that is dense in $[a,b]$. (Unfortunately that isn't necessarily true for sets other than intervals.) – B. S. Thomson Oct 28 '15 at 15:55
  • Thomson, I think that idea might be beyond what I'm supposed to know at this point. Is there any other way to explain it? I'm posting a link to my proof. If you could visit it, and let me know how to fix it so that it accurately reflects what you're saying, I would very much appreciate it. http://math.stackexchange.com/questions/1501940/showing-that-if-f-a-a-e-on-a-general-measurable-function-it-is-not-necessari –  Oct 28 '15 at 16:07
1

Statement:$$f=g\text{ a.e. on }E$$ is the same statement as: $$f1_E=g1_E\text{ a.e.}$$

If $m(E)=0$ then this statement is true for any pair of continuous functions $f,g$.

But $m(E)=0$ does not imply that $E=\varnothing$.

If $f(x)\neq g(x)$ for some $x\in E$ then the statement: $$f=g\text{ on }E$$ is false.

drhab
  • 151,093
  • But if $E$ is open and $m(E) = 0$, then yes it does imply that $E = \emptyset$, right? –  Oct 28 '15 at 15:41
  • Yes, that is correct. Unfortunately I have to go now. Later I will come back to have a second look. Cheers. – drhab Oct 28 '15 at 15:42
  • thanks. I need to edit my proof. –  Oct 28 '15 at 15:43
  • also, later, I still want you to explain to me why it works if $f=g$ a.e. on $[a,b]$. –  Oct 28 '15 at 15:44
  • If $h(x):=|f(x)-g(x)|>0$ for some $x\in[a,b]$ then the set $U:={y\in[a,b]\mid h(y)>0}$ is not-empty. Also $U$ is open in $[a,b]$ and consequently $m(U)>0$. Be cautious, though. This works for intervals, but not for every measurable set with positive measure. A counterexample is given by @saz in his answer to your next question. – drhab Oct 28 '15 at 17:50
1

The assertion is false for some $E$ with $m(E)> 0.$ For example it fails for $E=[0,1]\cup \{2\}.$ The assertion is false for any $E$ with $m(E)=0.$ Here just take $f=0,g=1$ on $E.$

zhw.
  • 105,693
0

Let $E = \{\ast\}$ be some set with a single element. Equip it with the trivial sigma-algebra and a measure $m$ that is constant $0$. Let $f: E \to \{1,2\}$ be the constant $1$ function. Let $g: E \to \{1, 2\}$ be the constant $2$ function. Trivially, $f$ and $g$ are continuous. Trivially, it holds $f=g$ almost everywhere (the only point $\ast$ does not count, because $m(\{\ast\}) = 0$). However, they are obviously not equal.

If $E$ has to be a subset of $\mathbb{R}$ equipped with the trace-sigma algebra of the standard Borel sigma algebra of $\mathbb{R}$, the argument still holds: simply take $E=\{0\}$, or pick any other real number.

Andrey Tyukin
  • 875