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Clearly, if the limit function \(f\) is integrable then the MCT becomes a special case of DCT. The advantage of Fatou’s lemma and MCT is that they are applicable even if \(f\) is not integrable and are often a good way to show that \(f\) is integrable.
**Lemma 13** (Fatou).: _Suppose \((f_{n})\) is a sequence of measurable functions with \(f_{n}\geq 0\). If \(f_{n}\to f\) a.e., then_
\[\int f\,dx\leq\liminf_{n\to\infty}\int f_{n}\,dx.\]
Before we prove Fatou’s lemma we observe the following points:
* The conclusion of Fatou’s lemma can be written as \[\int\liminf_{n\to\infty}f_{n}\,dx\leq\liminf_{n\to\infty}\int f_{n}\,dx.\]* In case \(\lim_{n\to\infty}\int f_{n}\) exists then Fatou’s lemma assures that \[\int\lim_{n\to\infty}f_{n}\,dx\leq\lim_{n\to\infty}\int f_{n}\,dx.\]
* \(f_{n}(x)\to f(x)\) a.e. means \(\lim_{n\to\infty}f_{n}(x)=f(x)\) for every \(x\in E\) and \(m(E^{c})=0\). Therefore \[\int f\,dx=\int_{E}f\,dx+\int_{E^{c}}f\,dx=\int_{E}f\,dx\] \[=\int f\chi_{E}\,dx\leq\liminf_{n\to\infty}\int f_{n}\chi_{E}\,dx= \liminf_{n\to\infty}\int_{E}f_{n}\,dx.\]
Proof
: Suppose \(g\) is a bounded function and supported on a set \(E\) of finite measure with \(0\leq g\leq f\). Set
\[g_{n}(x)=\min(g(x),f_{n}(x)).\]
Then
* \(g_{n}\) is measurable and supported on \(E\).
* \(g_{n}(x)\leq f_{n}(x)\) for all \(x\) and thus \(\int g_{n}\,dx\leq\int f_{n}\,dx\).
* \(g_{n}(x)\to g(x)\) a.e..
We can apply Theorem 116 (BCT) to \(g_{n}\) to conclude
\[\lim_{n\to\infty}\int g_{n}\,dx=\int\lim_{n\to\infty}g_{n}\,dx.\]
Therefore,
\[\int g\,dx=\lim_{n\to\infty}\int g_{n}\,dx\leq\lim_{n\to\infty}\int f_{n}\,dx\]
and \(\int g\,dx\leq\liminf_{n\to\infty}\int f_{n}\,dx\). Taking the supremum over all \(g\) with \(g\leq f\) gives the claimed inequality.
The following are some consequences of Fatou’s lemma.
**Corollary 23**.: _Suppose \(f\geq 0\) is a non-negative measurable function, and \(\{f_{n}\}\) a sequence of non-negative measurable functions with \(f_{n}(x)\leq f(x)\) and \(f_{n}(x)\to f(x)\) a.e.. Then_
Proof
: Since \(f_{n}(x)\leq f(x)\) a.e., we have \(\int f_{n}(x)\,dx\leq\int f(x)\,dx\) for all \(n\); hence
\[\limsup_{n\to\infty}\int f_{n}\,dx\leq\int f\,dx.\]
By Fatou’s lemma we also have
\[\int f\,dx\leq\liminf_{n\to\infty}\int f_{n}\,dx,\]
and combining these two inequalities proves the claim.
**Corollary 24** (Monotone Convergence Theorem).: _Suppose \((f_{n})\) is a sequence of non-negative measurable functions that satisfies_
\[\lim_{n\to\infty}f_{n}(x)=f(x)\text{ a.e }x.\quad\text{and}\quad f_{n}(x)\leq f _{n+1}\,(x)\,a.e\,x\text{ all }n\geq 1.\]
_Then_
\[\lim_{n\to\infty}\int f_{n}\,dx=\int f\,dx.\]
Proof
: Suppose \((f_{n})\) is a sequence of measurable functions that satisfies the given conditions. Since \((f_{n})\) is increasing, \(\lim_{n\to\infty}f_{n}(x)=\sup_{n}f_{n}(x)\) and
\[0\leq\int f_{n}\,dx\leq\int f_{n+1}\,dx\leq\int f\,dx.\]
Therefore \(\limsup_{n\to\infty}\int f_{n}\,dx\leq\int f\,dx\), and here again apply Fatou’s lemma to get the desired conclusion.
**Remark 66**.: MCT implies the linearity property of the integral of non-negative measurable functions. Because if \(f\) and \(g\) are two such functions we can choose two sequences of measurable functions \(\varphi_{n}\) and \(\psi_{n}\), both supported on a set of finite measure and \(\lim_{n\to\infty}\varphi_{n}=f\) a.e. and \(\lim_{n\to\infty}\psi_{n}=g\) a.e.. Then \(\varphi_{n}+\psi_{n}\) increases to \(f+g\) and by the MCT
\[\int(f+g)\,dx=\lim_{n\to\infty}\int(\varphi_{n}+\psi_{n})\,dx.\]
Now since both \(\varphi_{n}\) and \(\psi_{n}\) are bounded we can apply BCT to interchange limit and integral in the above equality to conclude
\[\int(f+g)\,dx=\int f\,dx+\int g\,dx.\]
**Corollary 25** (Beppo-Levi Theorem).: _Consider an infinite series of the form \(\sum_{k=1}^{\infty}a_{k}(x)\) where \(a_{k}(x)\geq 0\) and is measurable for every \(k\geq 1\). Then_
\[\int\sum_{k=1}^{\infty}a_{k}(x)\,dx=\sum_{k=1}^{\infty}\int a_{k}(x)\,dx.\]
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\[0\leq\int(1-\chi_{B_{N}})f\,dx=\int\chi_{B_{N}^{c}}f\,dx=\int_{B_{N}^{c}}f\,dx<\epsilon.\qed\]
**Theorem 118** (Dominated Convergence Theorem).: _Suppose \((f_{n})\) is a sequence of measurable functions such that \(f_{n}(x)\to f(x)\) a.e. as \(n\to\infty\). If_
\[|f_{n}(x)|\leq g(x)\,\text{ where }\,g\,\text{ is integrable},\]
_then_
\[\lim_{n\to\infty}\int f_{n}\,dx=\int f\,dx.\]
Proof
: For each \(N\) define \(E_{N}=\{x:\ |x|\leq N,\ g(x)\leq N\}\). Given \(\epsilon>0\), by the Lemma 14 and \(g\) being integrable there exists some \(N\) such that \(\int_{E_{N}^{c}}g\,dx<\epsilon\). Moreover, the function \(f\,\chi_{E_{N}}\leq N\) and is supported on a set of finite measure. Thus we can apply BCT to say
\[\int_{E_{N}}\,|\,f_{n}-f\,|\,\ dx<\epsilon\quad\text{for all large }\,\,n.\]
Hence
\[\int|f_{n}-f|\,dx =\int_{E_{N}}|f_{n}-f|\,dx+\int_{E_{N}^{c}}|f_{n}-f|\,dx\] \[\leq\int_{E_{N}}|f_{n}-f|\,dx+2\int_{E_{N}^{c}}g\,dx\] \[\leq\epsilon+2\epsilon.\qed\]
**Remark 67**.: It is possible to weaken the hypothesis of DCT to \(|f_{n}(x)|\leq g(x)\) a.e. and that \(f_{n}\to f\) a.e.. See also Exercise 10 below for another version.
### Exercises
1. Prove that if \(a,b\in\mathbb{R}\) and \(a 2. Prove that if \(f\) and \(g\) are measurable, then so is \(cf\) and \(fg\) where \(c\) is real. 3. Let \(\phi\) be continuous on \(\mathbb{R}\) and let \(f\) be finite a.e. in \(E\subset\mathbb{R}^{d}\), so that in particular \(\phi\circ f\) is defined a.e. in \(E\). 1. Show that \(\phi(f)\) is measurable if \(f\) is.
2. Show that \(|f|,|f|^{p}\) for \(p>0,\)and \(e^{cf}\) are measurable if \(f\) is.
3. Give an example of a function \(f\) which is not measurable but \(|f|\) is measurable.
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