Chapter 6: The Monotone Convergence Theorem
- Page ID
- 208226
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)A very useful result for nonnegative delta measurable functions is the monotone convergence theorem. This is a precursor to many convergence theorems which we shall prove later.
Let \(\{f_n\}_{n\in \mathbb{N}}\) be an increasing sequence of nonnegative Lebesgue delta measurable functions \(f_n: \mathbb{T}\to [0, \infty]\) that converges to some nonnegative Lebesgue delta measurable function \(f: \mathbb{T}\to [0, \infty]\). Then
\[ \begin{aligned} \int\limits_{\mathbb{T}}f\Delta\mu_\Delta=& \int\limits_{\mathbb{T}}\lim\limits_{n\to\infty}f_n \Delta\mu_\Delta\\ =& \lim\limits_{n\to\infty}\int\limits_{\mathbb{T}}f_n\Delta \mu_\Delta. \end{aligned} \notag\]
Proof
We have
\[ f_n\leq f_{n+1}\leq f,\quad n\in \mathbb{N}. \notag\]
Then
\[ \begin{aligned} \int\limits_{\mathbb{T}}f_n \Delta\mu_\Delta\leq& \int\limits_{\mathbb{T}}f_{n+1}\Delta\mu_\Delta\\ \leq& \int\limits_{\mathbb{T}}f\Delta \mu_\Delta,\quad n\in \mathbb{N}. \end{aligned} \notag\]
Taking the limit as \(n\to\infty\), we have
\[ \begin{aligned} \lim\limits_{n\to\infty}\int\limits_{\mathbb{T}}f_n\Delta\mu_\Delta=&\sup\limits_{n\in \mathbb{N}}\left(\int\limits_{\mathbb{T}}f_n\Delta\mu_\Delta\right)\\ \leq& \int\limits_{\mathbb{T}}f\Delta\mu_\Delta. \end{aligned} \notag\]
Now, we need to prove the inverse inequality. Let
\[ \phi(t)=\sum\limits_{j=1}^m c_j \mathbf{1}_{E_j}(t),\quad t\in \mathbb{T}, \notag\]
where \(\{E_j\}_{j=1}^m\) are pairwise disjoint and \(c_j\), \(j\in \{1, \ldots, m\}\), are real constants, and
\[ 0\leq \phi\leq f. \notag\]
Fix \(\alpha\in (0, 1)\) arbitrarily and define
\[ B_n=\{t\in \mathbb{T}: f_n(t)>\alpha \phi(t)\},\quad n\in \mathbb{N}. \notag\]
The sets \(B_n\), \(n\in \mathbb{N}\), are Lebesgue delta measurable and since the sequence \(\{f_n\}_{n\in \mathbb{N}}\) is increasing, we get
\[ B_n\subset B_{n+1},\quad n\in \mathbb{N}. \notag\]
Moreover, we have
\[ \mathbb{T}= \bigcup\limits_{n=1}^\infty B_n \notag\]
because
\[ \begin{aligned} f_n(t)\to& f(t)\\ >& \alpha \phi(t)\quad \mbox{as}\quad n\to \infty,\quad t\in \mathbb{T}. \end{aligned} \notag\]
Since
\[ \begin{aligned} \alpha \phi \mathbf{1}_{B_n}\leq& f_n \mathbf{1}_{B_n}\\ \leq& f_n,\quad n\in \mathbb{N}, \end{aligned} \notag\]
by integrating over \(\mathbb{T}\), we get \begin{equation} \label{50} \alpha I_{B_n}(\phi)\leq \int\limits_{\mathbb{T}}f_n\Delta\mu_\Delta,\quad n\in \mathbb{N}. \end{equation} Since
\[ E_j\cap B_n\subset E_j \cap B_{n+1},\quad j, n\in \mathbb{N}, \notag\]
and
\[ \bigcup\limits_{n=1}^\infty(E_j \cap B_n)=E_j, \notag\]
we have the following limit
\[ \begin{aligned} \lim\limits_{n\to\infty}I_{B_n}(\phi)=& \lim\limits_{n\to \infty}\sum\limits_{j=1}^m c_j \mu_\Delta(E_j\cap B_n)\\ =& \sum\limits_{j=1}^n c_j \lim\limits_{n\to\infty}\mu_\Delta(E_j\cap B_n)\\ =& \sum\limits_{j=1}^m c_j \mu_\Delta(E_j)\\ =& I(\phi). \end{aligned} \notag\]
Hence, taking the limit as \(n\to\infty\) in \eqref{50}, we arrive at
\[ \alpha I(\phi)\leq \lim\limits_{n\to \infty} \int\limits_{\mathbb{T}}f_n\Delta\mu_\Delta \notag\]
for any arbitrary \(\alpha\in (0, 1)\). Taking the limit as \(\alpha\to 1\) then yields
\[ I(\phi)\leq \lim\limits_{n\to \infty}\int\limits_{\mathbb{T}}f_n\Delta \mu_\Delta. \notag\]
Now, taking the supremum over all simple functions \(\phi\) with \(0\leq \phi\leq f\), we find
\[ \int\limits_{\mathbb{T}}f\Delta \mu_\Delta\leq \lim\limits_{n\to \infty}\int\limits_{\mathbb{T}}f_n\Delta\mu_\Delta. \notag\]
Putting the two inequalities together, we obtain the desired result.