Take two continuous functions $f,g$ ( for simplicity say $f,g:\mathbb{R}\to \mathbb{R}$).
If we know $$f\leq g~~ \text{almost everywhere (w.r.t the Lebesgue measure).}$$
Then does it follow that
$$ f(t) \leq g(t) ~~\text{for all $t\in\mathbb{R}$}.$$
Does anyone have a counter-example or a proof of this statement?
The reason I hoped this to be true is from the continuity of $f,g$ if : $g(s) < f(s)$ for some $s$ then there will be a small ball around $s$ where for each $t$ in such a ball $g(t)<f(t)$ - which is a contradiction. I struggled to prove this though.
All non-empty open sets have, indeed, strictly positive Lebesgue measure and $\{x\in\Bbb R\,:\, g(x)<f(x)\}$ is open.