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?

  • 3
    $\begingroup$ 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 $X \in \mathcal{A}$. $\endgroup$ Mar 14, 2021 at 8:17

1 Answer 1


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 $(X,\mathcal{T})$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .