# Does compactness imply measurability?

Let $$(\mathbb{R}^n,\mathcal{L})$$ be a measurable space, where $$\mathcal{L}$$ is the Lebesgue $$\sigma$$-algebra. Then,

1. $$X\subset\mathbb{R}^n$$ compact $$\Rightarrow X\in\mathcal{L}$$

Let $$(\mathbb{R}^n,\mathcal{B})$$ be a measurable space, where $$\mathcal{B}$$ is the Borel $$\sigma$$-algebra. Then,

1. $$X\subset\mathbb{R}^n$$ compact $$\Rightarrow X\in\mathcal{B}$$

Are (1) & (2) true statements? If we replace compact for closed, are (1) & (2) true?

Bonus question: When we say that a set $$X\subset\mathbb{R}^n$$ is Lebesgue (or Borel) measurable, what we mean is that $$X\in\mathcal{L}$$ (or $$X\in\mathcal{B}$$), correct?

• Every closed set is the complement of an open set. With that in mind, it's probably best for you to review the definition of a $\sigma$-algebra. It will also be useful that the borel $\sigma$-algebra contains all the open sets by definition. As for the bonus question, you're exactly right. We say a set $X$ is "$\mathcal{A}$ measurable" for a $\sigma$-algebra $\mathcal{A}$ exactly when $X \in \mathcal{A}$. Mar 14, 2021 at 8:17

Yes to both, trivially, because closed sets (being the complement of open sets which are by definition in the Borel $$\sigma$$-algebra) are Borel and hence Lebesgue measurable.
And yes to the "bonus question", just as $$O$$ open is the same as $$O \in \mathcal{T}$$ for a topological space $$(X,\mathcal{T})$$.