Measurable sets and functions

In one sentence

Countable, measure zero and nowhere dense are three different kinds of small, no two of them imply each other, and this topic is mostly the business of keeping them apart.

Why the exam asks it

Because "negligible" is a single word in English and at least three different things in analysis. A candidate who has not separated them will accept "measure zero, so countable" without blinking, and the Cantor set — uncountable, measure zero — is a one-word refutation that the paper expects you to have ready.

The idea

Outer measure of $E \subseteq \mathbb{R}$: the infimum of $\sum |I_k|$ over all countable covers of $E$ by open intervals. It is defined for every set, and it is only countably subadditive, not additive — which is the whole reason measurable sets have to be singled out.

$E$ is Lebesgue measurable if for every $A$,

$$m^*(A) = m^*(A \cap E) + m^*(A \cap E^c).$$

That is Carathéodory's criterion: $E$ splits every set additively. The measurable sets form a $\sigma$-algebra, and $m^*$ restricted to it is countably additive. Lebesgue measure is translation invariant, and every interval has its length.

Borel against Lebesgue

The Borel $\sigma$-algebra is the smallest one containing the open sets. Every Borel set is Lebesgue measurable. The converse is false, and the cleanest argument is by counting.

The Cantor set $C$ has measure zero, so every one of its subsets has outer measure zero and is therefore measurable. $C$ has cardinality $2^{\aleph_0}$, so it has $2^{2^{\aleph_0}}$ subsets, all measurable. But there are only $2^{\aleph_0}$ Borel sets. So most measurable sets are not Borel — not by a small margin.

A non-measurable set exists too: Vitali's, built by choosing one representative from each coset of $\mathbb{Q}$ in $\mathbb{R}$. It needs the axiom of choice, and no explicit example is possible without it.

The Cantor set, and its fat cousin

Remove the open middle third of $[0,1]$, then the middle thirds of what remains, forever. At stage $n$ you remove $2^{n-1}$ intervals of length $3^{-n}$, so the total removed is

$$\sum_{n=1}^{\infty}\frac{2^{n-1}}{3^{n}} = 1 .$$

What remains has measure zero. It is also uncountable — its points are exactly the numbers with a base-3 expansion using only $0$ and $2$, which is a bijection with all binary sequences. And it is nowhere dense and compact.

Now run the same construction removing intervals of length $4^{-n}$ instead:

$$\sum_{n=1}^{\infty}\frac{2^{n-1}}{4^{n}} = \frac12 .$$

The result is the fat Cantor set: still nowhere dense, still containing no interval, and of measure $\tfrac12$. So nowhere dense does not imply measure zero, and a set can have positive measure with empty interior.

Measurable functions

$f$ is measurable if $\{x : f(x) > a\}$ is measurable for every real $a$. That single family suffices; preimages of all open sets, and then of all Borel sets, follow.

The closure properties are generous, and this is the point of the theory:

• Sums, products, scalar multiples of measurable functions are measurable.
• $\sup_n f_n$, $\inf_n f_n$, $\limsup_n f_n$, $\liminf_n f_n$ are measurable.
• A pointwise limit of measurable functions is measurable.

That last one is the contrast worth holding on to: a pointwise limit of continuous functions need not be continuous, and a pointwise limit of Riemann integrable functions need not be Riemann integrable, but measurability survives pointwise limits without any hypothesis at all.

Egorov: on a set of finite measure, pointwise a.e. convergence is uniform off a set of arbitrarily small measure. Lusin: a measurable function is continuous off a set of arbitrarily small measure. Both say measurable functions are nearly as good as the classical ones.

Where intuition breaks

"Measure zero, so countable." The Cantor set. The most productive single counterexample in this topic. Countable does imply measure zero; the converse is what fails.

"Nowhere dense, so measure zero." The fat Cantor set, measure $\tfrac12$. Tempting because the middle-thirds set has both properties and it is the only such set most people have seen.

"Dense, so positive measure." $\mathbb{Q}$ is dense in $\mathbb{R}$ and has measure zero. Density and measure are unrelated.

"Lebesgue measurable, so Borel." False by the counting argument above. The inclusion runs Borel $\subsetneq$ Lebesgue, and the gap is enormous.

"$|f|$ is measurable, so $f$ is." Let $E$ be non-measurable and set $f = 1$ on $E$, $f = -1$ off it. Then $|f| \equiv 1$ is as measurable as a function can be, while $\{f > 0\} = E$ is not. A converse assumed: measurable implies $|f|$ measurable, not the reverse.

"$f$ and $g$ measurable, so $g \circ f$ is measurable." False. What is true is that $g \circ f$ is measurable when $f$ is measurable and $g$ is continuous — the composition needs the outer function to pull Borel sets back to Borel sets, and a merely measurable $g$ pulls them back to Lebesgue sets, which is not enough. Note the order: continuous-after-measurable is fine, measurable-after-continuous is not.

"An uncountable set has positive measure." Cantor again. Cardinality and measure are independent, in both directions.

"Every subset of $\mathbb{R}$ is measurable." Vitali. It requires the axiom of choice, which is why you will never be shown one explicitly.

The exam's angle

1. Name which smallness is being claimed. Countable, measure zero, nowhere dense, meagre — then read the table. Almost every false option in this topic asserts one from another.
2. Reach for the Cantor set first, the fat Cantor set second. Between them they break most of what can be asked.
3. For composition questions, check which function is continuous. Continuous outside, measurable inside is fine; the other order is not.
4. For "is this function measurable", try to build it from limits. Sups, infs, limsups and pointwise limits all preserve measurability, so almost anything assembled that way is measurable, and the question is usually whether the pieces are.
5. When an option involves a non-measurable set, expect it to be there to break an "obvious" converse — most often the $|f|$ one.

The night before

• Countable $\Rightarrow$ measure zero. Neither converse, and nowhere dense implies nothing about measure.
• Cantor set: uncountable, measure zero, nowhere dense, compact. Removed lengths sum to $1$.
• Fat Cantor set: nowhere dense, measure $\tfrac12$, contains no interval.
• $\mathbb{Q}$: countable, measure zero, dense.
• Borel $\subsetneq$ Lebesgue, by a cardinality argument on subsets of the Cantor set.
• Non-measurable sets exist (Vitali) and need the axiom of choice.
• Measurability survives sups, infs, limsups and pointwise limits. Continuity does not.
• $|f|$ measurable does not give $f$ measurable.
• $g$ continuous and $f$ measurable $\Rightarrow$ $g\circ f$ measurable. Not the other order.
• Egorov: almost uniform convergence on finite measure. Lusin: nearly continuous.