If $f=g$ a.e. and $f$ and $g$ are continuous, then $f=g$ every where.

Question

Let $f,g:\mathbb R\longrightarrow \mathbb R$ two continuous functions. Is it true that if $f=g$ a.e. then $f=g$ every where ?

I would say yes. My argument would be the following one : Let $$\Omega =\{x\mid f(x)\neq g(x)\}.$$

So let $x\in \Omega $. Is it true that there is a sequence in $\mathbb R\backslash \Omega $ that converge to $x$ ? If yes, the claim is proved. But if yes, How can I prove this claim ? If no, how can I do ?

Does this result can be adapted to an abstracted measurable space $(X,\mu)$ ?

In an abstract measure space $(X,\mu)$, how can one talk about continuity if $X$ has no topology? — , Jun 24 '17 at 14:19
To elaborate on Jack's point, for a more general version you need to be working with something like a strictly positive measure. — Anthony Carapetis, Jun 24 '17 at 14:23

score 8 · Accepted Answer · answered Jun 24 '17 at 14:17

8

Since $f$ and $g$ are continuous, $\Omega$ is open, if it is not empty, its measure is not zero.

answered Jun 24 '17 at 14:17

Tsemo Aristide

87,475

yes ! very nice proof ! – user330587 Jun 24 '17 at 14:18
In fact, how you prove that $\Omega $ is open ? And the argument of a sequence of $(x_n)\subset \mathbb R\backslash \Omega $ that converge to $x$ is wrong ? – user330587 Jun 24 '17 at 14:20
2

because it is equal to $(f-g)^{-1}(\mathbb{R}-{0})$ and $\mathbb{R}-{0}$ is open. – Tsemo Aristide Jun 24 '17 at 14:22

If $f=g$ a.e. and $f$ and $g$ are continuous, then $f=g$ every where.

1 Answers1