KindiakMath

The Measure-Theoretic Trifecta

Any self-respecting discussion in measure theory cannot ignore the three pillars of convergence: the monotone convergence theorem, Fatou’s lemma, and the dominated convergence theorem.

These results aim to answer the question: given measurable functions f : \Omega \to \mathbb R, what conditions on \{f_n\} do we need in order to guarantee the following interchange of limits?

\displaystyle \int_{\Omega} \lim_{n \to \infty} f_n\, \mathrm d\mu = \lim_{n \to \infty} \int_{\Omega} f_n\, \mathrm d\mu.

Recall the monotone convergence theorem for real numbers.

Lemma 1. Let \{x_n\} be a non-decreasing sequence of real numbers that is bounded above. Then \{x_n\} converges to \sup\{x_n\}.

We want to prove an analogous result for measurable functions. Let (\Omega, \mathcal F, \mu) be any measure space.

Lemma 2. For measurable f, g : \Omega \to [0, \infty], if f \leq g, then

\displaystyle \int_{\Omega} f\, \mathrm d\mu \leq \int_{\Omega} g\, \mathrm d\mu.

Proof. Fix \epsilon > 0. Then there exists a simple function 0 \leq \varphi \leq f \leq g such that

\displaystyle \int_{\Omega} f\, \mathrm d\mu - \epsilon \leq \int_{\Omega} \varphi\, \mathrm d\mu \leq \int_{\Omega} g\, \mathrm d\mu.

Taking \epsilon \to 0^+ yields the desired result.

Theorem 1 (Monotone Convergence Theorem). Let \{f_n\} be a sequence of measurable functions f_n : \Omega \to [0, \infty]. Then the map f : \Omega \to \mathbb R defined by

\displaystyle f(\omega) := \sup_n f_n(\omega)

is measurable. Furthermore, if \{f_n\} is increasing in n (that is, f_i \leq f_j whenever i < j), then f_n \to f and

\displaystyle \int_{\Omega} f\, \mathrm d\mu = \lim_{n \to \infty} \int_{\Omega} f_n\, \mathrm d\mu.

Proof. For any a \in [0, \infty], we have [a, \infty] \in \frak{B}([0,\infty]). Since each f_n is measurable, f_n^{-1}([a, \infty]) \in \mathcal F. Therefore,

\displaystyle f^{-1}([a, \infty]) = \bigcap_{n = 1}^\infty f_n^{-1}([a, \infty]) \in \mathcal F.

Now if \{f_n\} is increasing in n, then for any \omega \in \Omega, f_n(\omega) \to f(\omega) by Lemma 1. By Lemma 2,

\displaystyle \lim_{n \to \infty} \int_{\Omega} f_n\, \mathrm d\mu \leq \lim_{n \to \infty} \int_{\Omega} f\, \mathrm d\mu = \int_{\Omega} f\, \mathrm d\mu.

Thus, we claim that the reverse inequality

\displaystyle \int_{\Omega} f\, \mathrm d\mu \leq \lim_{n \to \infty} \int_{\Omega} f_n\, \mathrm d\mu

holds. Firstly, assume \int_{\Omega} f\, \mathrm d\mu = \infty. We claim that the right hand side equals \infty. Fix N > 0. By definition, there exists a simple function \varphi = \sum_{i=1}^n a_i \cdot \mathbb I_{K_i}, a_i > 0, such that 0 \leq \varphi \leq f and

\displaystyle \sum_{i=1}^n a_i \cdot \mu(K_i) = \int_{\Omega} \varphi\, \mathrm d\mu > N.

Define \epsilon := \min\{a_1,\dots,a_n\}/2. so that \epsilon^{-1} > a_i > \epsilon for each i. Define K := \bigsqcup_{i=1}^n K_i. There are two cases: \mu(K) < \infty or \mu(K) = \infty.

Suppose \mu(K) < \infty. Let \delta > 0 be a constant to be tuned. Define

L_n := \{\omega \in K : f_n(\omega) > f(\omega) - \delta\}.

Due to the monotonicity of \{f_n\}, L_1 \subseteq L_2 \subseteq \dots and K = \bigcup_{n=1}^\infty L_n. By continuity of measures, \mu(L_n) \to \mu(K). Hence, there exists m \in \mathbb N such that \mu(L_m) > \mu(K) - \delta. Therefore,

\begin{aligned} \int_{L_m} \varphi\, \mathrm d\mu &= \int_{L_m} \sum_{i=1}^n a_i \cdot \mathbb I_{K_i}\, \mathrm d\mu \\ &= \sum_{i=1}^n a_i \cdot \mu (K_i \cap L_m) \\ &= \sum_{i=1}^n a_i \cdot \mu(K_i) - \sum_{i=1}^n a_i \cdot \mu(K_i \backslash L_m) \\ &> \sum_{i=1}^n a_i \cdot \mu(K_i) - \max\{a_1,\dots,a_n\} \cdot \sum_{i=1}^n \mu(K_i \backslash L_m) \\ &> N - \epsilon^{-1} \cdot \delta. \end{aligned}

By the construction of L_m,

\begin{aligned} \int_{L_m} f_m \, \mathrm d\mu + \delta \cdot \mu(L_m) &= \int_{L_m} (f_m + \delta) \, \mathrm d\mu \\ &> \int_{L_m} f \, \mathrm d\mu\\ & \geq \int_{L_m} \varphi \, \mathrm d\mu \\ &> N - \epsilon^{-1} \cdot \delta. \end{aligned}

Therefore,

\begin{aligned} \int_{\Omega} f_m\, \mathrm d\mu &\geq \int_{L_m} f_m \, \mathrm d\mu \\ &> N - \epsilon^{-1} \cdot \delta - \delta \cdot \mu(L_m) \\ &> N - \delta \cdot (\epsilon^{-1} + \mu(K)) > N - 1,\end{aligned}

where we set \delta := 1/(\epsilon^{-1}+\mu(K)). That way, \int_{\Omega} f_m\, \mathrm d\mu \to \infty, as required.

Now suppose \mu(K) = \infty. It suffices to assume \mu(K_1) = \infty without loss of generality. Define

M_n := \{\omega \in K_1 : f_n(\omega) \geq \epsilon\}.

Similar to the previous case, M_1 \subseteq M_2 \subseteq \dots and \bigcup_{n=1}^\infty M_n = K_1. Thus, there exists \ell \in \mathbb N such that \mu(M_{\ell}) > N/\epsilon. Hence,

\displaystyle \int_{\Omega} f_{\ell}\, \mathrm d\mu \geq \int_{M_{\ell}} f_{\ell}\, \mathrm d\mu \geq \int_{\Omega} \epsilon \cdot \mathbb I_{M_{\ell}} \, \mathrm d\mu = \epsilon \cdot \mu(M_{\ell}) > N.

That way, \int_{\Omega} f_{\ell}\, \mathrm d\mu \to \infty, as required.

Finally, we work on the case where \int_{\Omega} f\, \mathrm d\mu < \infty. Fix \delta > 0. By definition, there exists a simple function \varphi = \sum_{i=1}^n a_i \cdot \mathbb I_{K_i}, a_i > 0, such that 0 \leq \varphi \leq f and

\displaystyle \sum_{i=1}^n a_i \cdot \mu(K_i) = \int_{\Omega} \varphi\, \mathrm d\mu > \int_{\Omega} f\, \mathrm d\mu - \delta.

Define \epsilon := \min\{a_1,\dots,a_n\}/2. so that \epsilon^{-1} > a_i > \epsilon for each i. Define K := \bigsqcup_{i=1}^n K_i. Then

\displaystyle \mu(K) = \sum_{i=1}^n \mu(K_i) < \epsilon^{-1} \cdot \sum_{i=1}^n a_i \cdot \mu(K_i) \leq \epsilon^{-1} \cdot \int_{\Omega} f\, \mathrm d\mu < \infty.

Fix \eta > 0 to be tuned. Define

R_n := \{\omega \in K : f_n(\omega) > \varphi(\omega) - \eta\},

so that R_1 \subseteq R_2 \subseteq \dots and \bigcup_{n=1}^\infty R_n = K. Hence for any \theta > 0, there exists k \in \mathbb N such that \mu(R_k) > \mu(K) - \theta. Hence,

\begin{aligned} \int_{\Omega} f_k\, \mathrm d\mu &\geq \int_{R_k} f_k\, \mathrm d\mu \\ &\geq \int_{R_k} (\varphi - \eta)\, \mathrm d\mu \\ &= \int_{R_k} \varphi\, \mathrm d\mu - \eta \cdot \mu(R_k) \\ &> \int_{\Omega} \varphi\, \mathrm d\mu - \epsilon^{-1} \cdot \theta - \eta \cdot \mu(K). \end{aligned}

Choose \theta = \epsilon \cdot \delta and \eta := \delta/\mu(K) to obtain the inequality

\begin{aligned} \int_{\Omega} f_k\, \mathrm d\mu &> \int_{\Omega} \varphi\, \mathrm d\mu - 2\delta &> \int_{\Omega} f\, \mathrm d\mu - 3\delta. \end{aligned}

That way, \int_{\Omega} f_k\, \mathrm d\mu \to \int_{\Omega} f\, \mathrm d\mu, as required.

The monotone convergence theorem therefore tells us that if \{f_n\} is monotonically increasing in n, then

\displaystyle \lim_{n \to \infty} \int_{\Omega} f_n\, \mathrm d\mu = \int_{\Omega} \lim_{n \to \infty} f_n\, \mathrm d\mu.

What happens if we do not have such a nice feature like montonicity? While equality feels far-fetched, we do get a useful partial result.

Theorem 2 (Fatou’s Lemma). Let \{f_n\} be a sequence of measurable functions f_n : \Omega \to [0, \infty]. Then the map f : \Omega \to \mathbb R defined by

\displaystyle f(\omega) := \liminf_{n \to \infty} f_n(\omega) \equiv \lim_{n \to \infty} \left\{ \inf_{k \geq n} f_k(\omega) \right\}

is measurable. Furthermore,

\displaystyle \int_{\Omega} f\, \mathrm d\mu \leq \lim_{n \to \infty} \int_{\Omega} f_n\, \mathrm d\mu.

Proof. We know that f is measurable because

\displaystyle f^{-1}([a, \infty]) = \bigcup_{n = 1}^\infty \left( \bigcap_{k=n}^\infty f_k^{-1}([0, \infty]) \right).

Define \tilde f_n := \inf_{k \geq n} f_k, which is measurable. By construction, \{ \tilde f_n\} is monotonically increasing and \tilde f_n \leq f_n. By the monotone convergence theorem,

\displaystyle \int_{\Omega} \liminf_{n \to \infty} f_n\, \mathrm d\mu = \int_{\Omega} \lim_{n \to \infty} \tilde f_n\, \mathrm d\mu = \lim_{n \to \infty} \int_{\Omega}\tilde f_n\, \mathrm d\mu \leq \lim_{n \to \infty} \int_{\Omega} f_n\, \mathrm d\mu.

Remark 1. We usually abbreviate Fatou’s lemma using the inequality

\displaystyle \int_{\Omega} \liminf_{n \to \infty}f_n\, \mathrm d\mu \leq \liminf_{n \to \infty} \int_{\Omega} f_n\, \mathrm d\mu.

To ensure the correct inequality direction, the example f_n = \mathbb I_{[n,n+1)} yields 0 on the left-hand side and 1 on the right-hand side. Thus, the integral of the point-wise limit is usually more “stringent” than the limit of the integral.

In a sense, the monotone convergence theorem requires a “strictest” requirement on the sequence \{f_n\} of measurable functions in order to interchange limits. Fatou’s lemma, on the other hand, requires almost “no” requirement on \{f_n\}, at the price of losing equality. Do we have some “middle ground” between the two?

Yes we do, and it is called Lebesgue’s dominated convergence theorem. In fact, we require the underlying sequence to be integrable (which is usually what we are interested in) and have some form of “boundedness” in order to interchange limits. This theorem turns out to more often than not be the most useful out of the three convergence theorems.

Theorem 3 (Dominated Convergence Theorem). Let \{f_n\} be a sequence functions f_n : \Omega \to \mathbb R satisfying the following properties:

  • each f_n is measurable,
  • there exists a measurable function f : \Omega \to \mathbb R such that f_n \to f,
  • there exists an integrable function g : \Omega \to [0, \infty) such that |f_n| \leq g.

Then f_n, f are integrable, and

\displaystyle \int_{\Omega} f\, \mathrm d\mu = \lim_{n \to \infty} \int_{\Omega} f_n\, \mathrm d\mu.

Proof. We first suppose each f_n \geq 0, which implies that f \geq 0. Since |f_n| \leq g,

\displaystyle \int_{\Omega} f_n \, \mathrm d\mu \leq \int_{\Omega} g \, \mathrm d\mu < \infty.

Thus, each f_n is integrable. By Fatou’s lemma,

\displaystyle \int_{\Omega} f\, \mathrm d\mu = \int_{\Omega} \liminf_{n \to \infty} f_n\, \mathrm d\mu \leq \lim_{n \to \infty} \int_{\Omega} f_n\, \mathrm d\mu \leq \int_{\Omega} g\, \mathrm d\mu < \infty,

since g is integrable, so that f is integrable.

On the other hand, applying Fatou’s lemma to the nonnegative sequence g - f_n \to g - f, since g-f is integrable,

\displaystyle \int_{\Omega} (g-f)\, \mathrm d\mu \leq \lim_{n \to \infty} \int_{\Omega} (g-f_n)\, \mathrm d\mu.

By the linearity of integration and algebruh,

\begin{aligned} \lim_{n \to \infty} \int_{\Omega} f_n\, \mathrm d\mu &= \int_{\Omega} g\, \mathrm d\mu-\lim_{n \to \infty} \int_{\Omega} (g-f_n)\, \mathrm d\mu \\ &\leq \int_{\Omega} g\, \mathrm d\mu-\int_{\Omega} (g-f)\, \mathrm d\mu = \int_{\Omega} f\, \mathrm d\mu. \end{aligned}

For the general case, decompose f_n = f_n^+ - f_n^- and f = f^+ - f^-. Apply the nonnegative case to f_n^+ \to f^+ and f_n^- \to f^- respectively since |f_n^+| \leq g and |f_n^-| \leq g. Combine the integrals to yield the desired result.

There are other measure-theoretic tools that can help us make sense of probability, but we will leave them to the next post. Here, we established the trifecta of measure theory up and front, ready-to-use for any future use case.

—Joel Kindiak, 13 Jul 25, 1439H

,

Published by

joelkindiak

Leave a comment