Vtyjkyuk

Here's a problem from Cohn's Measure Theory.

[Cohn, 2.1.3] Let $f$ and $g$ be continuous real-valued functions on $mathbbR$. Show that if $f =g$ for $lambda$-almost every $x$, then $f = g$ everywhere.

Here $lambda$ denotes Lebesgue measure.

Here's what I do.

Let $h: mathbbR to mathbbR$ be a continuous function. Suppose that $h neq 0$. Then for some $x in mathbbR$, there is $n in mathbbN$ such that $|h(x)| > 2/n$. Then by continuity for some $delta > 0$, we may ensure that whenever $|y - x| < delta/2$, we have
$$
|h(y)| geq |h(x)| - |h(y) - h(x)| > 2/n - 1/n = 1/n.
$$
Hence, $lambda(h neq 0) geq lambda( > 1/n) geq delta > 0$. Taking the contrapositive, we just showed if $h$ is a continuous real-valued function on $mathbbR$, then
$lambdah neq 0 = 0$ implies $h = 0$ (everywhere). The result follows now by setting $h = f - g$, which is continuous.

Question. Is this basically the simplest way to do it? Other ways?

I would say that $x:h(x)ne0$ is an open set in $Bbb R$. (Since it is
the inverse image of the open set $Bbb Rsetminus0$ under the continuous
function $h$). Each nonempty open set contains a nonempty open interval,
which has positive Lebesgue measure. So if $h$ is continuous, and not identically
zero, then it cannot be almost everywhere zero (w.r.t. Lebesgue measure).

(For completeness, following Lord Shark the Unknown's advice.)

Let $h: mathbbR to mathbbR$ be a continuous function. Suppose that $h neq 0$. Thus, the set $h neq 0$ is nonempty and open, and hence has positive Lebesgue measure. Taking the contrapositive, if $h$ is a continuous real-valued function on $mathbbR$, then $lambdah neq 0 = 0$ implies $h = 0$ (everywhere). Now take $h = f - g$. The result follows.