Why $f$ restricted to the set of its discontinuities $E_0$ is measurable?

Question

I am trying to answer the following question:

Suppose a function $f$ has a measurable domain and is continuous except at a finite number of points. Is $f$ necessarily measurable?

And I found this answer online:

"Let $E$ denote the (Lebesgue) measurable domain of $f$ and define $$E_0 = \{x \in E| f \text{ is not continuous at } x \}$$

Since $E_0$ is finite, $m(E_0) = 0$ and $f$ is measurable on $E_0.$ By proposition $3,$ $f$ is measurable on $E \sim E_0$ as it is continuous on this set. We conclude from proposition $5(ii)$ that $f$ is measurable on $E.$"

My question is:

Why $f$ restricted to the set of its discontinuities $E_0$ is measurable? by what theorem, proposition or logical justification?

Answer 1

Since $E_0$ is finite and therefore discrete, the restriction of the $\sigma$-algebra of Borel sets to $E_0$ is $\mathcal P(E_0)$. Therefore, any function defined on $E_0$ is measurable.

Edit

If $(X,\mathcal A)$ is a measurable space (ie $X$ is a set and $\mathcal A$ is a $\sigma$-algebra on $X$) and $Y\subset X$ is a subset, then $Y$ is canonically a measurable space $(Y,\mathcal A|_Y)$ where the restriction of the $\sigma$-algebra $\mathcal A$ to the subset $Y$ is defined as : $$\mathcal A|_Y= \{Y\cap W:W\in\mathcal A\}$$ If $X$ is a topological space and $\mathcal A = \mathcal B(X)$ is its Borel $\sigma$-algebra, then $\mathcal A|_Y = \mathcal B(Y)$ (ie the restriction of a Borel algebra is the Borel algebra of the topological subspace).