The Borel sets and the Lebesgue measurable sets

There is a relationship between the Borel sigma-algebra and the Lebesgue measurable sets (another sigma-algebra).

In short: Every Borel set is Lebesgue measurable, but there are Lebesgue measurable sets that are not Borel. The Lebesgue $\sigma$-algebra is essentially the Borel $\sigma$-algebra, but with a crucial "patch" applied to fix a logical annoyance regarding sets of size zero.

Here is exactly how they relate and why we need both.

1. The Borel $\sigma$-Algebra

The Borel $\sigma$-algebra, denoted $\mathcal{B}(\mathbb{R})$, is built from the ground up using the natural topology of the real number line.

• How it is built: You start with all open intervals (like $(0,1)$ or $(2.5,5)$). Then, you apply the three rules of a $\sigma$-algebra (complements and countable unions). You keep combining, complementing, and unioning these intervals to infinity.
• What it contains: The resulting collection, $\mathcal{B}(\mathbb{R})$, contains open sets, closed sets, points, and incredibly complex infinite unions of these.
• The philosophy: If a set can be logically constructed by taking limits of standard intervals, it is a Borel set. For almost every practical purpose in probability and statistics, the Borel $\sigma$-algebra is all you will ever need.

2. The Problem with Borel: The "Dust" of Measure Zero

We define the Lebesgue measure (let's call it $m$) as the standard measure of length. For an interval, $m([a,b])=b-a$.

Sometimes, a set has a length of exactly zero. For example, a single point has measure zero. A countable set of points (like all rational numbers) also has a total measure of zero. We can think of these measure-zero sets as mathematical "dust."

Now, logic dictates a basic rule of subsets: If a set has a total length of zero, then any subset of it should also have a length of zero. (If a pile of dust weighs zero pounds, half of that pile must also weigh zero pounds).

Here is where the Borel $\sigma$-algebra fails: There are Borel sets of measure zero that contain subsets that are not Borel sets. Through the lens of the Borel "VIP club," you have a set with length $0$ inside the club, but some of its subsets are arbitrarily kicked out of the club, meaning you aren't allowed to measure them at all. This is annoying.

3. The Lebesgue Sets: The "Completed" VIP Club

The Lebesgue measurable sets, denoted $\mathcal{L}(\mathbb{R})$, fix this exact problem. This process is called completion.

To get the Lebesgue $\sigma$-algebra, you do the following:

1. Take the entire Borel $\sigma$-algebra $\mathcal{B}(\mathbb{R})$.
2. Find every set that has a measure of $0$.
3. Force every single subset of those measure-zero sets into the club, and declare their measure to also be $0$.

Summary of the Relationship

Mathematically, the relationship is a strict hierarchy:

• $\mathcal{B}(\mathbb{R})$ (Borel): The pure, logically constructed sets built from intervals.
• $\mathcal{L}(\mathbb{R})$ (Lebesgue): The Borel sets plus all the subsets of measure-zero dust.
• $\mathcal{P}(\mathbb{R})$ (Power Set): Every possible subset. We still cannot measure this whole thing, because the chaotic, paradox-inducing sets (like the Vitali set) live here, outside of both $\mathcal{B}$ and $\mathcal{L}$.

A fun fact about their sizes: Even though the Lebesgue club just looks like the Borel club with some zero-length dust swept into it, that dust is infinitely vast. The number of Borel sets is the same as the number of real numbers (the continuum, $\mathfrak{c}$). But the number of Lebesgue measurable sets is drastically larger—it is $2^{\mathfrak{c}}$, which is the same size as the entire, unmeasurable power set $\mathcal{P}(\mathbb{R})$.