So I am trying to prove Luzin's Theorem, and I've gotten to
Lemma: If $f: \mathbb{R} \to \mathbb{R}$ is measurable, then for each $\epsilon > 0$, and for each measurable set $D$ of finite measure, there exists a measurable set $E$ with $m(E) < \epsilon$ such that the restriction of $f$ to $D \setminus E$ is continuous.
However, $D$ only has finite measure, and I am trying to prove it for $\mathbb{R}$. So I was wondering if this proof works:
Let $D_n = [-n, n]$. Apply the lemma to each $D_n$ and find a measurable set $E_n$ such that $m(E_n) < \epsilon/2^n$ and $f$ is only discontinuous on a subset of $E_n$. Then since $\mathbb{R} = \cup_{n = 1}^{\infty} D_n$, $f$ (as a function on $\mathbb{R}$) can only be discontinuous on $\cup_{n = 1}^{\infty} E_n$, which is a set of measure less than $m(E_1) + m(E_2) + ... = \epsilon(1/2 + 1/4 + ...) = \epsilon$. QED.
I am worried about the part where $f$ can only be discontinuous on $\cup_{n=1}^{\infty}E_n$. Is this a valid argument? Thank you!