Measure Theory

Problem: Show that the Lebesgue measure of the Cantor set is 0.

Solution:

Step 1: The Cantor set $C$ is constructed by repeatedly removing middle thirds from $[0,1]$.

Step 2: At stage $n$, we remove $2^{n-1}$ intervals, each of length $3^{-n}$.

Step 3: Total length removed:

Step 4: Since $[0,1]$ has Lebesgue measure 1 and we removed a total of 1, the remaining set $C$ has measure:

Result: The Cantor set is uncountable (it has cardinality $2^{\aleph_0}$) but has Lebesgue measure 0. This demonstrates that "size" (measure) and cardinality are fundamentally different notions, and motivates the need for σ-algebras to handle such pathological sets.

Problem: Compute $\lim_{n \to \infty} \int_0^\infty \left(1 + \frac{x}{n}\right)^{-n} e^{-x/2} \, dx$.

Solution:

Step 1: Identify the pointwise limit. For each fixed $x \geq 0$:

So $f_n(x) = \left(1 + \frac{x}{n}\right)^{-n} e^{-x/2} \to e^{-x} \cdot e^{-x/2} = e^{-3x/2}$.

Step 2: Verify monotonicity. For fixed $x$, $\left(1 + \frac{x}{n}\right)^{-n}$ is increasing in $n$ (since $(1+t/n)^n$ is increasing). Therefore $f_n(x)$ is increasing in $n$.

Step 3: Apply the Monotone Convergence Theorem:

Step 4: Compute the integral:

Result: $\lim_{n \to \infty} \int_0^\infty \left(1 + \frac{x}{n}\right)^{-n} e^{-x/2} \, dx = \frac{2}{3}$

The MCT allowed us to swap limit and integral because the sequence was monotonically increasing.

Problem: Let $f_n = n \cdot \mathbf{1}_{(0, 1/n]}$ on $([0,1], \mathcal{B}, \lambda)$. Compute $\liminf \int f_n \, d\lambda$ and $\int \liminf f_n \, d\lambda$.

Solution:

Step 1: Compute $\int f_n \, d\lambda = n \cdot \lambda((0, 1/n]) = n \cdot \frac{1}{n} = 1$ for all $n$.

So $\liminf_{n \to \infty} \int f_n \, d\lambda = 1$.

Step 2: Find the pointwise limit. For any $x > 0$, there exists $N$ such that $x > 1/N$, so $f_n(x) = 0$ for all $n \geq N$. Thus $\lim_{n \to \infty} f_n(x) = 0$ for all $x > 0$.

Also $f_n(0) = 0$ for all $n$. So $\liminf f_n = 0$ pointwise.

Step 3: Compute $\int \liminf f_n \, d\lambda = \int 0 \, d\lambda = 0$.

Step 4: Fatou's Lemma gives:

Result: Fatou's inequality is strict: $0 < 1$. The functions $f_n$ escape to infinity near 0 (they "spike" higher but over smaller intervals), so the integral of the limit (0) is strictly less than the limit of the integrals (1). This is why Fatou gives an inequality, not an equality.

Problem: Compute $\lim_{n \to \infty} \int_0^1 \frac{n x^{n-1}}{1 + x^n} \, dx$.

Solution:

Step 1: Let $f_n(x) = \frac{nx^{n-1}}{1+x^n}$. Find the pointwise limit.

For $x \in [0, 1)$: $x^n \to 0$, so $f_n(x) \to \frac{0}{1+0} = 0$.

For $x = 1$: $f_n(1) = \frac{n}{2} \to \infty$.

Since $\{1\}$ has measure 0, $f_n \to 0$ a.e.

Step 2: Find a dominating function. Note that:

By the mean value theorem, for $x \in [0,1)$:

But this bound grows with $n$. Instead, note:

This doesn't work. Let's try a different approach.

Actually, integrate directly:

Step 3: The integral is $\ln 2$ for all $n$, so the limit is $\ln 2$.

Step 4: Apply DCT: $f_n \to 0$ a.e., but the integrals converge to $\ln 2 \neq 0$. The DCT does NOT apply because we cannot find an integrable dominating function (the "mass" escapes to the point $x=1$, which has measure 0 but the function blows up there).

Result: $\lim_{n \to \infty} \int_0^1 f_n \, dx = \ln 2$, despite $f_n \to 0$ a.e. This demonstrates that pointwise convergence alone does not guarantee convergence of integrals—we need the domination condition.

Problem: Prove that $\lambda(\mathbb{Q} \cap [0,1]) = 0$.

Solution:

Step 1: $\mathbb{Q} \cap [0,1]$ is countable. Enumerate it as $\{q_1, q_2, q_3, \ldots\}$.

Step 2: For any $\epsilon > 0$, cover each $q_n$ with an interval of length $\epsilon/2^n$:

Step 3: By countable subadditivity of Lebesgue measure:

Step 4: Since $\mathbb{Q} \cap [0,1] \subseteq U$ and $\lambda(U) \leq \epsilon$ for all $\epsilon > 0$:

Result: The rationals in $[0,1]$ have Lebesgue measure zero. This means Lebesgue integration "ignores" the rationals—they form a set of measure zero, so modifying a function on the rationals does not change its Lebesgue integral.

Problem: Let $\mu$ be the counting measure on $(\mathbb{N}, 2^{\mathbb{N}})$. Verify that $\mu$ is a measure.

Solution:

Step 1: Check $\mu(\emptyset) = 0$. The counting measure of the empty set is 0 (no elements to count). ✓

Step 2: Check countable additivity. Let $A_1, A_2, \ldots$ be pairwise disjoint subsets of $\mathbb{N}$.

Since the $A_n$ are disjoint, each element of $\bigcup A_n$ belongs to exactly one $A_n$:

Step 3: Both sides are either finite and equal, or both are infinite. ✓

Result: The counting measure is a valid measure. Note that it is σ-finite (each $\{n\}$ has measure 1), and the counting measure is used to define $\ell^p$ spaces.

Problem: Compute $\int_0^1 \mathbf{1}_{\mathbb{Q}}(x) \, d\lambda$ (Lebesgue integral of the Dirichlet function).

Solution:

Step 1: The Dirichlet function is $f(x) = \mathbf{1}_{\mathbb{Q}}(x) = \begin{cases} 1 & x \in \mathbb{Q} \\ 0 & x \notin \mathbb{Q} \end{cases}$

Step 2: We showed that $\lambda(\mathbb{Q} \cap [0,1]) = 0$.

Step 3: Since $f = 0$ almost everywhere (a.e.) with respect to Lebesgue measure:

Step 4: Contrast with Riemann integral: The Riemann integral of $\mathbf{1}_{\mathbb{Q}}$ does not exist because in every subinterval of $[0,1]$, the function takes both values 0 and 1, so upper and lower Riemann sums never converge.

Result: The Lebesgue integral equals 0, while the Riemann integral does not exist. This demonstrates the power of Lebesgue integration: it can integrate highly discontinuous functions that Riemann integration cannot handle.

Problem: Compute $\int_0^1 \int_0^1 \frac{x - y}{(x + y)^3} \, dx \, dy$ and $\int_0^1 \int_0^1 \frac{x - y}{(x + y)^3} \, dy \, dx$. Do they give the same answer?

Solution:

Step 1: Note that $f(x,y) = \frac{x-y}{(x+y)^3}$ is NOT non-negative, so we need to check integrability.

Step 2: Compute the iterated integral $\int_0^1 \left(\int_0^1 \frac{x-y}{(x+y)^3} \, dx\right) dy$.

Inner integral: $\int_0^1 \frac{x-y}{(x+y)^3} \, dx$. Let $u = x + y$, $du = dx$:

Wait, let me redo this more carefully.

Step 3: Outer integral: $\int_0^1 \frac{-1}{(1+y)^2} \, dy = \left[\frac{1}{1+y}\right]_0^1 = \frac{1}{2} - 1 = -\frac{1}{2}$

Step 4: By symmetry ($f(x,y) = -f(y,x)$):

Result: The two iterated integrals give different values ($-1/2$ vs $1/2$). Fubini's theorem does NOT apply because $\int \int |f(x,y)| \, dx \, dy = \infty$ (the function is not integrable on the product space). This demonstrates the importance of checking the integrability condition in Fubini's theorem.

Problem: Let $P$ be the uniform distribution on $[0,1]$ (Lebesgue measure) and $Q$ be the distribution with density $q(x) = 2x$. Find the Radon-Nikodym derivative $\frac{dQ}{dP}$.

Solution:

Step 1: Check absolute continuity. If $P(A) = \lambda(A) = 0$, then $Q(A) = \int_A 2x \, dx = 0$. So $Q \ll P$.

Step 2: By the Radon-Nikodym theorem, there exists $\frac{dQ}{dP}$ such that $Q(A) = \int_A \frac{dQ}{dP} \, dP$.

Step 3: Since $P$ is Lebesgue measure and $Q$ has density $2x$ with respect to Lebesgue measure:

Step 4: Therefore $\frac{dQ}{dP}(x) = 2x$.

Verification: $\int_0^1 2x \, dx = [x^2]_0^1 = 1 = Q([0,1])$ ✓

Result: The Radon-Nikodym derivative $\frac{dQ}{dP} = 2x$ is the likelihood ratio. In statistics, this is the "score function" or "likelihood ratio" used in hypothesis testing and maximum likelihood estimation. The Kullback-Leibler divergence can be expressed as $D_{KL}(Q \| P) = \int \log\left(\frac{dQ}{dP}\right) dQ$.

Problem: Let $A_n$ be events in a probability space with $\sum_{n=1}^{\infty} P(A_n) < \infty$. Show that $P(\limsup A_n) = 0$.

Solution:

Step 1: Define $B = \limsup A_n = \bigcap_{n=1}^{\infty} \bigcup_{k=n}^{\infty} A_k$ (infinitely many $A_n$ occur).

Step 2: For each $N$, $\bigcup_{k=N}^{\infty} A_k$ is an event containing $B$.

Step 3: By countable subadditivity:

Step 4: Since $\sum P(A_n) < \infty$, the tail $\sum_{k=N}^{\infty} P(A_k) \to 0$ as $N \to \infty$.

Step 5: Therefore $P(B) \leq P\left(\bigcup_{k=N}^{\infty} A_k\right) \leq \sum_{k=N}^{\infty} P(A_k) \to 0$.

Result: $P(\limsup A_n) = 0$. This means "almost surely, only finitely many $A_n$ occur." The Borel-Cantelli lemma is fundamental in probability theory and connects measure theory to the study of random events.

Problem: State Egorov's theorem and explain its significance.

Solution:

Statement: Let $(\Omega, \mathcal{F}, \mu)$ be a finite measure space ($\mu(\Omega) < \infty$), and let $f_n \to f$ pointwise a.e. Then for every $\epsilon > 0$, there exists a measurable set $E \subseteq \Omega$ with $\mu(E^c) < \epsilon$ such that $f_n \to f$ uniformly on $E$.

Significance: Egorov's theorem bridges pointwise convergence and uniform convergence. It says that on a finite measure space, pointwise convergence is "almost uniform"—we can find a set of arbitrarily small complement on which convergence is uniform.

Example: For $f_n = n \cdot \mathbf{1}_{(0,1/n]}$ on $[0,1]$, $f_n \to 0$ pointwise but not uniformly on $[0,1]$. However, for any $\epsilon > 0$, take $E = [\epsilon, 1]$. Then $\mu(E^c) = \epsilon$ and $f_n = 0$ on $E$ for $n > 1/\epsilon$, so $f_n \to 0$ uniformly on $E$.

Result: Egorov's theorem is important because it shows that pointwise convergence on finite measure spaces is "close to" uniform convergence, which is stronger and more useful for analysis.

Problem: State the Lebesgue differentiation theorem and explain its significance.

Solution:

Statement: For $f \in L^1_{\text{loc}}(\mathbb{R})$ (locally integrable), for almost every $x \in \mathbb{R}$:

Equivalently: $\lim_{r \to 0} \frac{1}{|B(x,r)|} \int_{B(x,r)} f(y) \, dy = f(x)$ a.e.

Significance: This theorem says that the "average value" of $f$ over small balls converges to $f(x)$ for a.e. $x$. It justifies the informal idea that integrals "recover" the integrand.

Example: For $f = \mathbf{1}_{[0,1]}$, at $x = 0.5$:

At $x = 0$ (boundary): $\frac{1}{2h}\int_{-h}^{h} \mathbf{1}_{[0,1]}(t)dt = \frac{h}{2h} = 1/2 \neq f(0) = 1$

But $\{0\}$ has measure zero, so the theorem holds a.e.

Result: The Lebesgue differentiation theorem is fundamental in real analysis—it connects the integral and the function value, and is used in the study of Hardy-Littlewood maximal functions, singular integrals, and the theory of differentiation.

Problem: State Jensen's inequality and apply it to show that $E[\log X] \leq \log E[X]$ for positive random variable $X$.

Solution:

Statement: If $\phi$ is a convex function and $X$ is a random variable, then $\phi(E[X]) \leq E[\phi(X)]$.

Step 1: Let $\phi(x) = -\log x$ (which is convex for $x > 0$).

Step 2: Apply Jensen's inequality:

Step 3: Therefore $E[\log X] \leq \log E[X]$.

Step 4: This is the AM-GM inequality in expectation: the geometric mean is always less than or equal to the arithmetic mean.

Example: If $X$ takes values 1 and 3 with equal probability:

• $E[X] = 2$, $\log E[X] = \log 2 \approx 0.693$
• $E[\log X] = \frac{1}{2}(\log 1 + \log 3) = \frac{1}{2}\log 3 \approx 0.549$
• $0.549 \leq 0.693$ ✓

Result: Jensen's inequality is fundamental in information theory (KL divergence is non-negative) and in machine learning (variational inference bounds).

Problem: Explain the completion of a measure space and why it matters.

Solution:

Definition: Given a measure space $(\Omega, \mathcal{F}, \mu)$, the completion $(\Omega, \bar{\mathcal{F}}, \bar{\mu})$ is defined by:

and $\bar{\mu}(A \cup B) = \mu(A)$ for $B \subseteq N$ with $\mu(N) = 0$.

Step 1: The completion adds all subsets of $\mu$-null sets to $\mathcal{F}$. This ensures that subsets of measure-zero sets are measurable.

Step 2: The Lebesgue measure is the completion of the Borel measure on $(\mathbb{R}, \mathcal{B}(\mathbb{R}))$. The Borel σ-algebra is generated by open sets, but there are Borel sets that are not Lebesgue measurable. The Lebesgue σ-algebra includes all Borel sets plus all subsets of Borel null sets.

Step 3: Why it matters:

• Without completion, subsets of null sets might not be measurable
• The Lebesgue differentiation theorem requires completion to state "almost everywhere" precisely
• In probability, completing the probability space ensures that events that "almost surely" don't happen have probability 0

Step 4: Example: The Cantor set has Lebesgue measure 0. Any subset of the Cantor set has Lebesgue measure 0 in the completed measure space, even though it may not be a Borel set.

Result: Completion ensures that measure theory is "maximally inclusive"—all subsets of negligible sets are automatically measurable. This is important for rigorous statements about "almost everywhere" convergence and for working with completions of probability spaces.

Problem: Construct the product measure on $[0,1]^2$ from Lebesgue measure on $[0,1]$.

Solution:

Step 1: Start with the product σ-algebra $\mathcal{B}([0,1]) \otimes \mathcal{B}([0,1]) = \mathcal{B}([0,1]^2)$.

Step 2: Define the product measure $\mu \times \mu$ on rectangles: $(\mu \times \mu)(A \times B) = \mu(A) \cdot \mu(B)$ for $A, B \in \mathcal{B}([0,1])$.

Step 3: By Carathéodory's extension theorem, this extends uniquely to a measure on $\mathcal{B}([0,1]^2)$.

Step 4: The product measure on $[0,1]^2$ is just the 2-dimensional Lebesgue measure $\lambda_2$.

Step 5: Verification via Fubini: For any $A \in \mathcal{B}([0,1]^2)$:

where $A_x = \{y : (x,y) \in A\}$ is the $x$-section of $A$.

Result: The product of 1-dimensional Lebesgue measure with itself gives 2-dimensional Lebesgue measure. This generalizes: $\lambda_n = \lambda_1 \times \cdots \times \lambda_1$ ($n$ times) gives $n$-dimensional Lebesgue measure.

Problem: State Tonelli's theorem and explain how it differs from Fubini's theorem.

Solution:

Statement: Let $(X, \mathcal{A}, \mu)$ and $(Y, \mathcal{B}, \nu)$ be σ-finite measure spaces. If $f: X \times Y \to [0, \infty]$ is measurable (non-negative), then:

1. The function $x \mapsto \int_Y f(x,y) \, d\nu(y)$ is measurable on $X$
2. $\int_{X \times Y} f \, d(\mu \times \nu) = \int_X \left(\int_Y f(x,y) \, d\nu(y)\right) d\mu(x) = \int_Y \left(\int_X f(x,y) \, d\mu(x)\right) d\nu(y)$

Key difference from Fubini: Tonelli requires $f \geq 0$ (non-negative) but does NOT require integrability. Fubini requires integrability but allows signed functions. For non-negative functions, you can always swap the order of integration (Tonelli). For signed functions, you must check integrability first (Fubini).

Example: For $f(x,y) = e^{-xy} \sin(x)$ on $[0,\infty) \times [0,\infty)$:

• Tonelli applies to $|f|$ to check integrability
• If $\int \int |f| < \infty$, then Fubini applies and we can swap

Result: Tonelli's theorem is often used as a stepping stone: first apply Tonelli to $|f|$ to verify integrability, then apply Fubini to $f$ to swap integration order.