I want to show that if $f$ and $g$ are continuous real-valued functions on $\mathbb{R}$ such that $f=g$ $\lambda$-a.e. ($\lambda$ is the Lebesgue measure) then $f=g$ $\,$ everywhere.
I think that using the following fact is necessary:
FACT: If $N\subseteq \mathbb{R}$ is $\lambda$-null, then $N^c$ is dense in $\mathbb{R}$.
Since $f=g$ $\lambda$-a.e. then I know that $f(x)=g(x)$ for all $x\in \mathbb{R}\setminus N$, where $N$ is a $\lambda$-null set. I will have to show that $f(x)=g(x)$ for all $x\in N$ as well. How could I apply the fact above? Thank you
Note that if $\lambda(N)=0$ then $R\setminus N$ is everywhere dense in $R$. Indeed, if assume the contrary then there will be $(a,b)$ such that $(R \setminus N)\cap (a,b)=\emptyset$. This means that $N$ contains the set $(a,b)$ and hence $N$ is not of $\lambda$-measure zero.
Since $f(x)=g(x)$ for $x \in R \setminus N$, $\lambda(N)=0$ and $R\setminus N$ is everywhere dense in $R$, for $y \in N$ you can choose $(x_k)$ from $R \setminus N$ such that $\lim_{n \to \infty}x_k=y$. Then we get $f(y)=\lim_{n \to \infty}f(x_k)=\lim_{n \to \infty}g(x_k)=g(y)$.