Show that $f(x) =g(x) $ for all $x \in \mathbb{R}$ if and only if $f=g $ $\lambda$-almost-everywhere

Question

Let $f, g:\mathbb{R} \rightarrow \mathbb{R}$ be two continous functions.

Show that $f(x) =g(x) $ for all $x \in \mathbb{R}$

if and only if

$f=g $ $\lambda$-almost-everywhere, , where $\lambda$ denotes Lebesgue measure.

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So I have shown the first implication which is quite simple

"$\implies$": Suppose $f(x) =g(x) $ for all $x \in \mathbb{R}$ $\implies f=g \implies f=g$ $\lambda$-almost-everywhere specifically.

It's the other implications that gives me some trouble. Since f and g are assumed to be continous I assume I have to use that property for something but I don't give see what. Any help would be appreciated, thanks in advance!

Answer 1

$HINT$

If a subset $E$ of the real line has measure zero,then $E^c$ is dense.

Use this an the sequential property of continuity.

Answer 2

If $f=g$ $\lambda$-almost everywhere, then $f-g=0$ $\lambda$-almost everywhere and $f-g$ is still continuous.

Assume there exists a point where $f-g\neq 0$. We can choose sequence of points going to that point, such that on every of them $f-g=0$. From continuity, $f-g$ has to be $0$ also in that point, contradiction.