Prob. 1, Chap. 3, in Royden's REAL ANALYSIS: If continuous functions $f$ and $g$ agree a.e. on $[a,b]$, then $f=g$ on $[a,b]$

Question

Here is Prob. 1, Chap. 3, in the book Real Analysis by H. L. Royden and P. M. Fitzpatrick, 4th edition:

Suppose $f$ and $g$ are continuous functions on $[a, b]$. Show that if $f = g$ a.e. on $[a, b]$, then, in fact, $f = g$ on $[a, b]$. Is a similar assertion true if $[a, b]$ is replaced by a general measurable set?

My Attempt:

Let $E_0$ be a subset of $[a, b]$ such that $m^* \left( E_0 \right) = 0$ and $f(x) = g(x)$ for all $x \in [a, b] \setminus E_0$.

What next? How to proceed from here?

Answer 1

The set $[0,1]\cup\{2\}$ is measurable. Let $f(x) = x$ for $x$ in this set, and let $g(x)= x$ for $x\in[0,1]$ and $g(2)=3.$ Then $f$ and $g$ agree almost everywhere on their domain, and both are continuous on that domain, but they don't agree everywhere.

Answer 2

Let $E=\{x\in [a,b]:f(x)\ne g(x)\}$. Since $E$ has measure $0$, for each $x\in [a,b]$, there exists a sequence $\{x_n\}$ in $[a,b]\setminus E$ such that $x_n\to x$. (Why?)

Since $f$ and $g$ are continuous, $$0 = \lim_{n\to\infty}[f(x_n)-g(x_n)] = f(x)-g(x).$$