Exercises — Measure theory — Lebesgue measure (intro)

(a) TRUE. By (B3), ${\mu }^{\ast }([a,b])=b-a$, so $5-2=3$. (b) TRUE. The empty set is covered by an interval of length $\epsilon$ for any $\epsilon >0$, so ${\mu }^{\ast }(\varnothing )\le \epsilon \to 0$, and outer measure is never negative. (c) FALSE. Every $\ell (I_n)=b-a\ge 0$, so every cover-sum is $\ge 0$, so the infimum is $\ge 0$. Outer measure is always $\ge 0$. (d) FALSE. By (Tool A), the whole point (and strength) of ${\mu }^{\ast }$ is that it is defined for every subset $A\subseteq \mathbb{R}$.

${\mu }^{\ast }(A)={\mu }^{\ast }(A\cap E)+{\mu }^{\ast }(A\cap E^c).$ In words: $E$ cleanly cuts every set $A$ into an inside-piece $A\cap E$ and an outside-piece $A\cap E^c$ whose outer measures add exactly — no leak (this is Tool C). See Sigma-algebra for why the collection of such $E$ is closed under complements and countable unions.

WHAT we do: cover each of the 4 points $p_k$ by a tiny interval $(p_k-\frac{\delta }{2},p_k+\frac{\delta }{2})$ of length $\delta$. WHY: each point needs only one interval; by (B2) the outer measure is at most the total cover length, and finitely many tiny intervals give a total we control. ${\mu }^{\ast }(\{1,4,9,16\})\le 4\delta \stackrel{\delta \to 0}{\to }0.$ Since outer measure is $\ge 0$, we get ${\mu }^{\ast }(\{1,4,9,16\})=0$. Any finite set has measure $0$.

Key observation: $A$ is countable (it is a list indexed by $n$). See Countable vs uncountable. WHAT we do: cover the $n$-th point $\frac{1}{n}$ by an interval of length $\frac{\epsilon }{2^n}$. WHY the $\epsilon /2^n$ choice: the lengths then form a geometric series whose total is finite and controllable; by sub-additivity (B2), ${\mu }^{\ast }(A)\le {\sum }_{n=1}^{\infty }\frac{\epsilon }{2^n}=\epsilon \cdot \frac{1/2}{1-1/2}=\epsilon \stackrel{\epsilon \to 0}{\to }0.$ So ${\mu }^{\ast }(A)=0$. This works even though the points pile up (accumulate) at $0$ — accumulation does not create length.

What the figure shows (below): the real line with the points $1,\frac{1}{2},\frac{1}{3},\dots$ (magenta dots) crowding toward $0$, each wrapped in a violet covering interval whose width halves each step ($\epsilon /2^n$). The picture makes visible that even as the dots pile up near $0$, the total violet length is the geometric sum $\epsilon$ — arbitrarily small. Read it as: "infinitely many dots, finite (shrinkable) total ink."

Upper bound ("$\le$"): cover with $(-\epsilon ,1+\epsilon )$ and $(3-\epsilon ,4+\epsilon )$; by (B2) ${\mu }^{\ast }\le (1+2\epsilon )+(1+2\epsilon )=2+4\epsilon \to 2$. So ${\mu }^{\ast }\le 2$. Lower bound ("$\ge$") — stated correctly in terms of ${\mu }^{\ast }$ and the split test: each of $[0,1]$ and $[3,4]$ is an interval, hence measurable (Tool C), and disjoint. Apply the Carathéodory split test of $[0,1]$ to the test set $A=[0,1]\cup [3,4]$, using $E=[0,1]$: ${\mu }^{\ast }(A)={\mu }^{\ast }(A\cap E)+{\mu }^{\ast }(A\cap E^c)={\mu }^{\ast }([0,1])+{\mu }^{\ast }([3,4])=1+1=2,$ where the two pieces are evaluated by (B3). Notice this is a statement about ${\mu }^{\ast }$ throughout — the equality (not just "$\le$") is exactly what measurability of $[0,1]$ delivers; sub-additivity (B2) alone would only give "$\le$". Combining both bounds: ${\mu }^{\ast }([0,1]\cup [3,4])=2$. $■$

The subtlety: a cover being dense (its intervals reach into every part of $[0,1]$) does not mean it fills $[0,1]$. The covering intervals overlap and leave gaps, and their total length — which by (B2) is all ${\mu }^{\ast }$ ever cares about — can be tiny even while their union is topologically spread out. Concrete, quantified picture: enumerate the rationals $q_1,q_2,\dots$ and put an interval of length $\epsilon /2^n$ around $q_n$. The union $U={\bigcup }_n{I}_n$ is a dense open set, yet by (B2) ${\mu }^{\ast }(U)\le {\sum }_n\frac{\epsilon }{2^n}=\epsilon .$ Take $\epsilon =0.01$: $U$ contains every rational in $[0,1]$, so its intervals poke into every sub-interval, yet ${\mu }^{\ast }(U)\le 0.01$. Its complement in $[0,1]$ is a nowhere-dense set (Tool D) of measure $\ge 0.99$ — a "fat" cousin of the Cantor set. Density is topology; measure is size; they are genuinely independent.

What the figure shows (below): the segment $[0,1]$ with a scatter of rationals (violet), each capped by a magenta interval of length $\epsilon /2^n$. Even though magenta appears near every part of the line (dense), the magenta total width is $\le \epsilon$; the untouched peach gaps are the positive-measure leftover. Read it as: "reaches everywhere, but occupies almost nothing."

WHAT we do: at stage $n$ of the construction, $C$ is contained in $2^n$ closed intervals each of length $3^{-n}$. WHY that covers $C$: the middle-third removal only ever deletes points, so $C$ stays inside the surviving intervals at every stage. Total length at stage $n$ (by (B2)): ${\mu }^{\ast }(C)\le 2^n\cdot 3^{-n}={\left(\frac{2}{3}\right)}^n\stackrel{n\to \infty }{\to }0.$ So ${\mu }^{\ast }(C)=0$. Why shocking: $C$ is uncountable (it bijects to infinite $\{0,2\}$-strings via base-3 digits), yet has measure $0$. Compare Example 3 of the parent: $[0,1]$ is uncountable with measure $1$. Conclusion: "uncountable" and "positive measure" are independent — you can have either without the other. See Cantor set and Countable vs uncountable.

(a) The $n$-th interval has length $2\cdot \frac{\epsilon }{2^{n+1}}=\frac{\epsilon }{2^n}$. By countable sub-additivity (B2), ${\mu }^{\ast }(U_{\epsilon })\le {\sum }_{n=1}^{\infty }\frac{\epsilon }{2^n}=\epsilon .$ (b) With $\epsilon =\frac{1}{2}$: ${\mu }^{\ast }(U_{1/2})\le \frac{1}{2}$. Now $[0,1]=(U_{1/2}\cap [0,1])\cup K$ where $K=[0,1]\setminus U_{1/2}$. Using ${\mu }^{\ast }([0,1])=1$ from (B3) and sub-additivity (B2), $1={\mu }^{\ast }([0,1])\le {\mu }^{\ast }(U_{1/2}\cap [0,1])+{\mu }^{\ast }(K)\le \frac{1}{2}+{\mu }^{\ast }(K),$ so ${\mu }^{\ast }(K)\ge \frac{1}{2}$. It is nonempty (its measure is positive). ✔ (c) $U_{1/2}$ is open and dense (it contains every rational in $[0,1]$) yet has measure $\le \frac{1}{2}$; the leftover $K$ is nowhere dense (Tool D: its closure contains no interval, since every interval meets a rational and hence meets $U_{1/2}$) yet has measure $\ge \frac{1}{2}$. So "topologically big" (dense open) and "measure-big" can point in opposite directions. This is the rigorous engine behind Problem 3.1.

WHAT we must check (Tool C): for every test set $T$, ${\mu }^{\ast }(T)={\mu }^{\ast }(T\cap A)+{\mu }^{\ast }(T\cap A^c).$ Step 1 — "$\le$" is free by sub-additivity (B2): ${\mu }^{\ast }(T)={\mu }^{\ast }((T\cap A)\cup (T\cap A^c))\le {\mu }^{\ast }(T\cap A)+{\mu }^{\ast }(T\cap A^c)$. Step 2 — "$\ge$": By monotonicity (B1), since $T\cap A\subseteq A$, we get ${\mu }^{\ast }(T\cap A)\le {\mu }^{\ast }(A)=0$, so ${\mu }^{\ast }(T\cap A)=0$. Also $T\cap A^c\subseteq T$ gives ${\mu }^{\ast }(T\cap A^c)\le {\mu }^{\ast }(T)$ by (B1). Adding: ${\mu }^{\ast }(T\cap A)+{\mu }^{\ast }(T\cap A^c)\le 0+{\mu }^{\ast }(T)={\mu }^{\ast }(T).$ Combine: both inequalities give equality, so $A$ is measurable. For a subset $B\subseteq A$: ${\mu }^{\ast }(B)\le {\mu }^{\ast }(A)=0$ by (B1), so ${\mu }^{\ast }(B)=0$, and by the same argument $B$ is measurable. $■$ Payoff: this is why the Cantor set (measure $0$) and all its subsets are automatically measurable — no leaks possible when there's no length to leak.

The blocker: the Vitali set $V$. Build it using the Axiom of Choice: partition $[0,1]$ by the equivalence relation $x\sim y⟺x-y\in \mathbb{Q}$, and pick one representative from each class — that choice needs the Axiom of Choice — to form $V$. The contradiction: enumerate the rationals in $[-1,1]$ as $q_1,q_2,\dots$ and form the translates $V_k=V+q_k$. Three facts:

1. The $V_k$ are pairwise disjoint, so their union is a genuine ${⨆}_k{V}_k$ (Tool C's symbol): two representatives differing by a rational would be in the same class, contradiction.
2. $[0,1]\subseteq {⨆}_k{V}_k\subseteq [-1,2]$.
3. Translation invariance — the wishlist property that "sliding a set does not change its length", $\mu (A+t)=\mu (A)$ — would force $\mu (V_k)=\mu (V)=:c$ for all $k$, since each $V_k$ is $V$ slid by $q_k$. Now apply countable additivity (the disjoint-union property $\mu ({⨆}_k{E}_k)={\sum }_k\mu (E_k)$) together with (B3): $1=\mu ([0,1])\le {\sum }_{k=1}^{\infty }c\le \mu ([-1,2])=3.$

• If $c=0$: the middle sum is $0$, contradicting $\ge 1$.
• If $c>0$: the middle sum is $+\infty$, contradicting $\le 3$. No value of $c$ works. So $V$ cannot be assigned a measure consistent with translation invariance and countable additivity simultaneously. Hence not every set is measurable — we must restrict to the $\sigma$-algebra $\mathcal{M}$.

What the figure shows (below): the interval $[-1,2]$ with $[0,1]$ marked (dashed), and several coloured rows of dots — each row is a translate $V+q_k$ of the same "seed" set $V$, shifted left/right by a rational. The rows are disjoint (no dot repeats) yet together they blanket $[0,1]$. The caption line reads off the trap: identical-measure disjoint copies sum to either $0$ or $\infty$, never fitting between $1$ and $3$.

No — the diagnosis is subtler. Sub-additivity (B2) says ${\mu }^{\ast }\left({\bigcup }_k{V}_k\right)\le {\sum }_k{\mu }^{\ast }(V_k)$, an inequality that always holds and creates no contradiction by itself. The problem is that additivity fails: for the disjoint $V_k$ we cannot have ${\mu }^{\ast }\left({⨆}_k{V}_k\right)={\sum }_k{\mu }^{\ast }(V_k)$ together with translation invariance. It is the loss of the equality (additivity), not any single inequality being "wrong", that is fatal. Equivalently: $V$ fails Carathéodory's exact-split test (Tool C) — it "leaks" measure. That leak is permitted for outer measure (sub-additive is all we ever promised on all sets, via (B2)); it is forbidden on the measurable sets, which is exactly why $V\notin \mathcal{M}$.

Any finite or countable set has outer measure ::: $0$ (cover the $n$-th point with length $\epsilon /2^n$; total $\le \epsilon \to 0$ by (B2)). A set can be uncountable and still have measure zero — example ::: the Cantor set, ${\mu }^{\ast }=\lim (2/3{)}^n=0$. "Dense" and "positive measure" are ::: independent — an open dense set can have measure $\le \epsilon$. If ${\mu }^{\ast }(A)=0$ then $A$ is ::: measurable, and so is every subset of $A$ (monotonicity (B1) forces the split). The object that proves not every set is measurable ::: the Vitali set, built via the Axiom of Choice; it breaks translation invariance + countable additivity.