12 — Uniform Integrability, Product Measures, and Fubini

Lecture 12 has two major components:

Finishing uniform integrability (UI): Proving that UI + a.s. convergence implies 𝐿 1 L 1 convergence, revisiting the key tail argument, and checking the conditions for UI.

Product measures & Fubini/Tonelli: Constructing the product measure from rectangles, proving uniqueness, and finishing with examples illustrating when Fubini works and when it fails.

All diagrams in the handwritten notes—including the rectangle-splitting picture on page 2 and the UI tail-vanishing picture on page 1—are incorporated conceptually.

We continue with:

A complete proof that UI + a.s. convergence ⇒ $L^1$ convergence,
The construction of product measures on $(\Omega_1 \times \Omega_2, \mathcal{F}_1 \otimes \mathcal{F}_2)$,
Tonelli (non-negative functions) and Fubini (integrable functions) theorems,
Counterexamples where Fubini fails because $\int \vert f\vert = \infty$.

1. Definition of Uniform Integrability

For a sequence ${X_n}\subset L^1$, $\phi(M) = \sup_{n \ge 1} \mathbb{E}\big[ \vert X_n\vert \mathbf{1}_{\{\vert X_n\vert >M\}} \big].$

\[\{X_n\} \text{ is UI } \iff \phi(M) \to 0\quad (M\to\infty).\]

2. Alternative Characterization of UI

${X_n}$ is UI iff:

$\sup_n \mathbb{E}\vert X_n\vert < \infty$,
For all $\varepsilon>0$ there exists $\delta>0$ such that
whenever $P(A)<\delta$, $\sup_n \mathbb{E}\big(\vert X_n\vert \mathbf{1}_A\big) < \varepsilon.$

This captures the idea that the sequence cannot concentrate large mass on small-probability sets (page 1).

3. Basic Applications of UI

As stated on page 1:

If ${X_n}$ and ${Y_m}$ are UI, then
${X_n + Y_m}$ is UI.
If $X_n \to X$ in $L^1$, then ${X_n}$ is UI.
If $\mathbb{E}\vert X_n\vert <\infty$ for all $n$, $\mathbb{E}\vert X\vert <\infty$, and $\mathbb{E}\vert X_n - X\vert \to 0,$ then ${X_n}$ is UI (follows from the two previous items).

4. Main Theorem: UI + a.s. Convergence ⇒ $L^1$ Convergence

Theorem.
If

$X_n \to X$ a.s.,
${X_n}$ is UI,

then: $\mathbb{E}\vert X_n - X\vert \to 0.$

Proof (as in pages 1–2)

Step 1. Show $X\in L^1$.

By Fatou: $\mathbb{E}\vert X\vert = \mathbb{E}\big[\liminf \vert X_n\vert \big] \le \liminf \mathbb{E}\vert X_n\vert \le \sup_n \mathbb{E}\vert X_n\vert < \infty.$

Step 2. Reduce the problem.

Define $Z_n = X_n - X$.
Then UI of ${X_n}$ and a.s. convergence imply UI of ${Z_n}$ (using closure of UI under addition/subtraction, page 1–2).

Thus it suffices to show: $X_n \to 0 \text{ a.s. and } \{X_n\}\text{ UI } \quad\Longrightarrow\quad \mathbb{E}\vert X_n\vert \to 0.$

Step 3. Tail decomposition (page 1 diagram)

Fix $M$:

\[\vert X_n\vert = \vert X_n\vert \mathbf{1}_{\{\vert X_n\vert \le M\}} + \vert X_n\vert \mathbf{1}_{\{\vert X_n\vert >M\}}.\]

Integrate: $\mathbb{E}\vert X_n\vert \le \mathbb{E}\big( \vert X_n\vert \mathbf{1}_{\{\vert X_n\vert \le M\}} \big) + \sup_k \mathbb{E}\big(\vert X_k\vert \mathbf{1}_{\{\vert X_k\vert >M\}}\big).$

Second term → $0$ as $M\to\infty$ by UI.
First term → $0$ as $n\to\infty$ for fixed $M$, by Bounded Convergence (since on ${\vert X_n\vert \le M}$ we have boundedness and $X_n\to 0$ a.s.)

Finally send $M\to\infty$.

So $\mathbb{E}\vert X_n\vert \to 0$.

Thus: $\boxed{ X_n\to X \text{ a.s. and UI } \Longrightarrow X_n\to X \text{ in }L^1. }$

5. Product Measures

Let
$(\Omega_1,\mathcal{F}_1,\mu_1)$ and $(\Omega_2,\mathcal{F}_2,\mu_2)$ be σ–finite.

Consider the collection of finite disjoint unions of measurable rectangles: $\mathcal{L} = \left\{ \bigcup_{i=1}^n (A_i \times B_i) :\, A_i\in\mathcal{F}_1,\, B_i\in\mathcal{F}_2,\, (A_i\times B_i)\cap(A_j\times B_j)=\varnothing\,(i\neq j) \right\}.$

Claim: $\mathcal{L}$ is an algebra (page 2).

Define: $\mu(A\times B)=\mu_1(A)\mu_2(B).$

Theorem (existence/uniqueness, page 2):
There exists a unique measure $\mu$ on
$\mathcal{F}_1\otimes \mathcal{F}_2=\sigma(\mathcal{L})$
whose value on rectangles is the product: $\mu(A\times B)=\mu_1(A)\mu_2(B).$

Key step of proof (page 2 diagram)

For disjoint rectangles decomposing $A \times B$: $\mathbf{1}_{A\times B} = \sum_{n=1}^\infty \mathbf{1}_{A_n}(\omega_1)\mathbf{1}_{B_n}(\omega_2).$

Integrate w.r.t. $\mu_1\otimes\mu_2$ and apply MCT to justify: $\mu(A\times B) = \sum_{n=1}^\infty \mu_1(A_n)\mu_2(B_n).$

6. Tonelli and Fubini Theorems

Let $\mu=\mu_1\otimes\mu_2$.

Tonelli (nonnegative $f$)

If $f\ge 0$ is measurable on $\Omega_1\times\Omega_2$, then both $x\mapsto \int_{\Omega_2} f(x,y)\, d\mu_2(y), \quad y\mapsto \int_{\Omega_1} f(x,y)\, d\mu_1(x)$ are measurable, and: $\int_{\Omega_1\times\Omega_2} f\, d\mu = \int_{\Omega_1}\!\left( \int_{\Omega_2} f\, d\mu_2 \right)d\mu_1 = \int_{\Omega_2}\!\left( \int_{\Omega_1} f\, d\mu_1 \right)d\mu_2.$

Fubini (integrable $f$)

If $\int \vert f\vert \, d\mu < \infty$, then:

Both iterated integrals exist and are finite,
And the same equality holds.

7. Example: Fubini Fails When $\int\vert f\vert =\infty$

From page 2:

Take
$\Omega_1=(0,1)$, $\Omega_2=(1,\infty)$, Lebesgue measures.

\[f(x,y)=e^{-xy}-2e^{-2xy}.\]

Then:

$\int_0^1\int_1^\infty f(x,y)\, dy\, dx > 0$,
$\int_1^\infty\int_0^1 f(x,y)\, dx\, dy < 0$.

The two iterated integrals disagree because: $\int_{\Omega_1\times\Omega_2} \vert f\vert \, d\mu = \infty,$ so Fubini does not apply (Tonelli also does not, because $f$ takes negative values).

8. Discrete Example with Counting Measure

From page 3:

Let
$\Omega_1=\Omega_2=\mathbb{N}$,
$\mathcal{F}_1=\mathcal{F}_2=2^{\mathbb{N}}$,
$\mu_1=\mu_2$ = counting measure.

Define $f(n,m)=a_{n,m}$ where:

$a_{i,i}=1$ for $i\ge 0$,
$a_{i+1,i}=-1$.

The matrix sketch on page 3 shows a diagonal of 1’s and a subdiagonal of −1’s.