Lebesgue-measurable or Borel-measurable

Question

In practice, does one always use Lebesgue-measurability? So do we always use Lebesgue-measurable instead of Borel-measurable on $\mathbb{R}^n$ because it is usually more convenient?

Answer 1

The Lebesgue measure is simply the closure of the Borel measure.

As an analogy, think about the number line. Basic operations give us the rational numbers, and that is all fine and dandy as far as basic algebra is concerned. But once we start thinking about things such as topology, we add in extra numbers called the irrational numbers to "close" the gaps in rational numbers.

With the above situation, closure refers to gaps which can be found by looking at Cauchy sequences (or supremums/infimums of subsets if you prefer). With measure, we start with a measure space $(A,M(A),m)$, where $m : M(A) \rightarrow [0,+\infty]$ is the measure function on the sets in $M(A) \subseteq A$.

The gaps of a measure space are considered to be any subset $X \subseteq A$ which can be contained in a measureable set of arbitrarily small size. (In other words: for all $\varepsilon > 0$, there exists $Y_\varepsilon \in M(A)$ with $X \subseteq Y$ and $m(Y_\varepsilon) \leq \varepsilon$). These are essentially the "missing" null sets from a measure space. Then the closure of $(A,M(A),m)$ is simply the smallest $\sigma$-algebra which contains both $M(A)$ and these missing null sets as measurable sets. (The function $m$ extends in a unique way).

So to summarize. There is no difference between Borel and Lebesgue measure in terms of what size various sets are, the Lebesgue measure just includes certain sets which the Borel measure misses out.