Probability with Measure

4.6 The dominated convergence theorem

We continue to work over a general measure space \((S,\Sigma ,m)\). Let \(f,g:S\to \R \) be measurable. We say that \(g\) dominates \(f\) if \(|f|\leq |g|\) almost everywhere. This concept is naturally connected to \(\Lone \) via the following lemma.

Proof: By part 1 of Lemma 4.2.2 we have \(\int _S |f|\,dm\leq \int _S|g|\,dm\), which completes the proof. (Exercise: Why don’t we use Theorem 4.5.3 here?)   ∎

We now present the second of our convergence theorems, the famous Lebesgue dominated convergence theorem - an extremely powerful tool in both the theory and applications of modern analysis.

The monotone convergence theorem is the basis of the interaction of Lebesgue integrals with limits, but it has the disadvantage that it only applies to monotone sequences of functions. The dominated convergence theorem does not require monotonicity but does require that all functions involved are dominated by some \(g\in \Lone \). For this reason \(g\) is often known as a dominating function for the \((f_n)\). The monotone and dominated convergence theorems are often known for short as the MCT and DCT. The proof of the DCT appears in Section 4.6.1. This completes our development of the Lebesgue integral.

We’re now in a position to use Lebesgue integration for doing calculations with integrals. We’ll do so in Section 4.7, which will include examples of both the MCT and DCT.

4.6.1 Proof of the DCT \((\star )\)

The proof of the DCT is off-syllabus, although we might cover it in lectures if we have time. We’ll begin with a famous lemma before we give the main proof. It is called Fatou’s lemma after the French mathematician and astronomer Pierre Fatou (1878-1929). It tells us how \(\liminf \) and \(\int \) interact. Understanding this is they key step for moving from monotone seqences of functions; non-monotone sequences of functions might not have pointwise limits but they do always have pointwise \(\liminf \)s.

Proof: Define \(g_{n} = \inf _{k \geq n}f_{k}\). Then \((g_{n})\) is an increasing sequence which converges to \(\li f_{n}\). Now as \(f_{l} \geq \inf _{k \geq n}f_{k}\) for all \(l \geq n\), Lemma 4.2.2(1)) we have that for all \(l \geq n\)

\[ \int _{S} f_{l}\, dm \geq \int _{S} \inf _{k \geq n}f_{k}\, dm,\]

and so

\[\inf _{l \geq n}\int _{S} f_{l}\, dm \geq \int _{S} \inf _{k \geq n}f_{k}\, dm.\]

Take limits on both sides of this last inequality and then apply the monotone convergence theorem (on the right hand side) to obtain

\begin{align*} \li \int _{S}f_{n}\,dm & \geq \lim _{n \rightarrow \infty } \int _{S} \inf _{k \geq n}f_{k}\, dm \\ & = \int _{S}\lim _{n \rightarrow \infty } \inf _{k \geq n}f_{k}\, dm \\ & = \int _{S}\li f_{n}\,dm \end{align*} as required.   ∎

Note that we do not require \((f_{n})\) to be a bounded sequence, so \(\li f_{n}\) should be interpreted as an extended measurable function, as discussed at the end of Chapter 2. The corresponding result for \(\limsup \), in which case the inequality is reversed, is known as the reverse Fatou lemma and can be found as Exercise 4.12.

Proof of Theorem 4.6.2: Note that we didn’t assume explicitly that \(f_{n}\in \Lone \), because this fact follows immediately from the first assumption and Lemma 4.6.1. For the same reason as in the proof of Theorem 4.3.1, we may assume without loss of generality that \(f_n\to f\) pointwise and that \(|f_n|\leq |g|\) pointwise. Thus \(f\) is measurable by Theorem 3.1.5. We may further assume without loss of generality that \(|f_n|\leq g\), by using \(|g|\) in place of \(g\) and noting that \(g\in \Lone \iff |g|\in \Lone \).

Since \(f_{n}\to f\) pointwise, \(|f_{n}|\to |f|\) pointwise. By Fatou’s lemma (Lemma 4.6.3) and monotonicity from Theorem 4.5.3, we have

\begin{align*} \int _{S}|f| \,dm & = \int _{S} \li |f_{n}|\,dm \\ & \leq \li \int _{S}|f_{n}| \,dm \\ & \leq \int _{S} g \,dm < \infty , \end{align*} so \(f\in \Lone \).

For all \(\nN \), since \(|f_n|\leq g\) we have \(g + f_{n} \geq 0\), so by Fatou’s lemma

\begin{equation} \label {eq:dct_fatou_1} \int _{S} \li (g + f_{n}) \,dm \leq \li \int _{S}(g + f_{n}) \,dm. \end{equation}

As pointwise limits we have \(\liminf _n (g + f_{n}) = g + \lim _{n}f_{n} = g+f\) so by linearity from Theorem 4.5.3 we have

\begin{align*} \int _{S} \li (g + f_{n}) \,dm &= \int _S g + \li f_n\,dm = \int _S g\,dm + \int _S \li f_n\, dm, \\ \li \int _{S}(g + f_{n}) \,dm &= \int _{S}g \,dm + \li \int _{S}f_{n} \,dm \end{align*} Putting these two equations in (4.19) gives that

\begin{equation} \label {eq:dct_1} \int _{S} f dm \leq \li \int _{S}f_{n}dm. \end{equation}

Next, we repeat the argument with \(g + f_{n}\) replaced by \(g - f_{n}\). Note that \(|f_n|\leq g\) gives that \(g-f_n\) is also non-negative for all \(\nN \). The result of doing so is

\[ -\int _{S} f dm \leq \li \left (-\int _{S}f_{n}dm\right )\]

which from Lemma 2.2.3 rearranges to

\begin{equation} \label {eq:dct_2} \int _{S} f dm \geq \ls \int _{S}f_{n}dm \end{equation}

Combining (4.20) and (4.21) we see that

\begin{equation} \label {eq:dct_3} \ls \int _{S}f_{n}dm \leq \int _{S} f dm \leq \li \int _{S}f_{n}dm. \end{equation}

Recall that \(\liminf _n a_n\leq \limsup _n a_n\) for any sequence, so in fact all three terms in (4.22) are equal. It now follows from Lemma 2.2.2 that \(\int _{S} f dm = \lim _{n}\int _{S} f_{n}dm.\)   ∎