📂Measure Theory

Proof of the Dominated Convergence Theorem

Theorem ¹

For a measurable set $E \in \mathcal{M}$ and $g \in \mathcal{L}^{1} (E)$, suppose a sequence of measurable functions $\left\{ f_{n} \right\}$ satisfies $|f_{n}| \le g$ almost everywhere on $E$. If $\displaystyle f = \lim_{n \to \infty} f_{n}$ almost everywhere on $E$, then $f \in \mathcal{L}^{1}(E)$ and $$ \lim_{ n \to \infty} \int_{E} f_{n} (x) dm = \int_{E} f dm $$

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• $f,g \in \mathcal{L}^{1} (E)$ means that $f$ and $g$ are Lebesgue integrable functions.

Explanation

Compared with the Monotone Convergence Theorem, the condition $f_{n} \nearrow f$ has been dropped, and it is no longer even required that $f_{n} \ge 0$.

Interestingly, although a $g$ that can ‘dominate’ $\left\{ f_{n} \right\}$ is needed, $g$ does not actually appear in the result.

Proof

Part 1.

Let us show that $f \in \mathcal{L}^{1}(E) $.

Since $|f_{n}| \le g$ on $E$, we have $-g(x) \le f_{n} \le g(x)$ for all $x \in E$. Rearranging, $$0 \le f_{n} (x) + g(x) \le 2 g(x)$$ and as $n \to \infty $, $$0 \le f (x) + g(x) \le 2 g(x)$$ so $$(f+g) \in \mathcal{L}^{1}(E)$$ Meanwhile, since $f = (f + g ) + ( -g)$ and $\mathcal{L}^{1}(E)$ is a vector space, $f \in \mathcal{L}^{1}(E)$.

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Part 2.

Assume that $f_{n} \ge 0$.

Fatou’s Lemma: For a sequence $\left\{ f_{n} \right\}$ of nonnegative measurable functions, $$\displaystyle \int_{E} f dm \le \liminf_{n \to \infty} \int_{E} f_{n} dm $$

By the assumption and Fatou’s Lemma, $$\displaystyle \int_{E} f dm \le \liminf_{n \to \infty} \int_{E} f_{n} dm $$ and it suffices to show that $\displaystyle \limsup_{n \to \infty} \int_{E} f_{n} dm \le \int_{E} f dm $.

Applying Fatou’s Lemma again to $g-f_{n}$, $$\displaystyle \int_{E} \lim_{n \to \infty} (g - f_{n}) dm \le \liminf_{n \to \infty} \int_{E} (g - f_{n} ) dm $$ Here, since $f, g \ge 0$, the left-hand side is $$\displaystyle \int_{E} \lim_{n \to \infty} (g - f_{n}) dm =\int_{E} g dm - \int_{E} f dm$$ and the right-hand side is $$ \begin{align*} & \liminf_{n \to \infty} \int_{E} (g - f_{n} ) dm \\ =& \liminf_{n \to \infty} \left( \int_{E} g dm - \int_{E} f_{n} dm \right) \\ =& \int_{E} g dm - \limsup_{n \to \infty} \int_{E} f_{n} dm \end{align*} $$ Rearranging, $$ \int_{E} g dm - \int_{E} f dm \le \int_{E} g dm - \limsup_{n \to \infty} \int_{E} f_{n} dm $$ Since $g \in \mathcal{L}^{1} (E)$, we have $\displaystyle \int_{E} g dm < \infty$, so it can be canceled from both sides, and rearranging the signs we obtain the following. $$\limsup_{n \to \infty} \int_{E} f_{n} dm \le \int_{E} f dm$$

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Part 3.

Let us generalize to the case where $f_{n} \ge 0$ does not hold. Defining $h_{n} := f_{n} + g$, we have $h_{n} \ge 0$, so the process carried out in Part 2 can be repeated.

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