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

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?

• Let $g = f|_{E_0} : E_0 \to \mathbb{R}$. The pre-image $g^{-1}(S) \subseteq E_0$ of any measurable $S \subseteq \mathbb{R}$ is a finite set, and hence measurable. Thus $g$ is measurable by definition.
– joeb
Oct 29, 2021 at 13:38
• @joeb why are you sure that the preimage is a finite set? and why we needed $S$ to be measurable in $\mathbb R$ Oct 29, 2021 at 14:23
• In this case, $E_0$ is a finite set, so all its subsets are also finite. Oct 29, 2021 at 15:46

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.
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).
• Sorry, I did not get what you said here "the restriction of the $\sigma$-algebra of Borel sets to $E_0$ is $\mathcal P(E_0)$" could you provide more details please? Oct 29, 2021 at 13:57