Chapter 10: Convergence Theorems
- Page ID
- 208230
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)We can improve the Monotone Convergence Theorem to general Lebesgue delta measurable functions.
Let \(\{f_n\}_{n\in \mathbb{N}}\). be a sequence of Lebesgue delta measurable functions \(f_n: \mathbb{T}\to \widetilde{\mathbb{R}}=[-\infty, \infty]\) that is pointwise increasing to some function \(f: \mathbb{T}\to \mathbb{R}\) and \(f_n\to f\) as \(n\to\infty\) \(\mu_\Delta\)-a.e. Suppose further that the set of Lebesgue delta integrals
\[ \left\{ \int\limits_{\mathbb{T}}f_n\Delta\mu_\Delta\right\}_{n\in \mathbb{N}} \notag\]
is bounded. Then \(f\in \mathcal{L}^1_\Delta(\mathbb{T})\) and
\[ \begin{aligned} \int\limits_{\mathbb{T}}f\Delta\mu_\Delta=& \int\limits_{\mathbb{T}}\lim\limits_{n\to\infty} f_n\Delta \mu_\Delta\\ =& \lim\limits_{n\to\infty} \int\limits_{\mathbb{T}}f_n\Delta \mu_\Delta. \end{aligned} \notag\]
Proof
Define
\[ g_n=f_n-f_1,\quad n\in \mathbb{N}. \notag\]
We have that \(g_n: \mathbb{T}\to [0, \infty]\) and the sequence \(\{g_n\}_{n\in \mathbb{N}}\) is increasing. Then we can apply the Monotone Convergence Theorem to get
\[ \begin{aligned} \lim\limits_{n\to\infty} \int\limits_{\mathbb{T}}g_n\Delta\mu_\Delta=& \lim\limits_{n\to\infty}\int\limits_{\mathbb{T}}(f_n-f_1)\Delta\mu_\Delta\\ =& \int\limits_{\mathbb{T}}\lim\limits_{n\to\infty}(f_n-f_1)\Delta\mu_\Delta\\ =& \int\limits_{\mathbb{T}}(f-f_1)\Delta\mu_\Delta. \end{aligned} \notag\]
Since
\[ \int\limits_{\mathbb{T}}f_1\Delta\mu_\Delta \notag\]
is finite, we can use algebra to get
\[ \lim\limits_{n\to\infty}\int\limits_{\mathbb{T}}f_n\Delta\mu_\Delta -\int\limits_{\mathbb{T}}f_1\Delta\mu_\Delta=\int\limits_{\mathbb{T}}f\Delta\mu_\Delta-\int\limits_{\mathbb{T}}f_1 \Delta\mu_\Delta, \notag\]
or
\[ \lim\limits_{n\to\infty} \int\limits_{\mathbb{T}}f_n \Delta\mu_\Delta=\int\limits_{\mathbb{T}}f\Delta\mu_\Delta. \notag\]
Finally, since the sequence of Lebesgue delta integrals
\[ \int\limits_{\mathbb{T}}f_n\Delta \mu_\Delta \notag\]
is bounded above and since limits preserve weak inequalities, the resulting limit
\[ \int\limits_{\mathbb{T}}f\Delta\mu_\Delta \notag\]
must be finite and so \(f\in \mathcal{L}^1_\Delta(\mathbb{T})\). This completes the proof.
Let \(f:\mathbb{T}\to \widetilde{\mathbb{R}}\) be a Lebesgue delta measurable function. For any \(\epsilon>0\) there exists a \(\delta>0\) such that for any set \(E\subset \mathbb{M}(m^*)\) with \(\mu_\Delta(E)<\delta\), we must have
\[ \int\limits_E|f|\Delta\mu_\Delta<\epsilon. \notag\]
Proof
Note that \(|f|\) is a Lebesgue delta measurable function. Define the sets
\[ E_n=\{t\in \mathbb{T}: |f(t)|\leq n\},\quad n\in \mathbb{N}, \notag\]
and the functions
\[ g_n=|f|\mathbf{1}_{E_n},\quad n\in \mathbb{N}. \notag\]
Note that \(\{g_n\}_{n\in \mathbb{N}}\) is pointwise increasing to \(|f|\). By the Monotone Convergence Theorem, it follows that there exists an index \(N\in \mathbb{N}\) such that
\[ \begin{aligned} \left|\int\limits_{\mathbb{T}}g_n\Delta\mu_\Delta -\int\limits_{\mathbb{T}}|f|\Delta\mu_\Delta\right|=& \int\limits_{\mathbb{T}}(|f|-g_n)\Delta\mu_\Delta\\ <& \frac{\epsilon}{2}. \end{aligned} \notag\]
For any \(E\subset \mathbb{M}(m^*)\) with \(\mu_\Delta(E)<\infty\), since \(|f|\geq g_N\) on \(\mathbb{T}\) and \(g_N\leq N\) on \(E\), we get
\[ \begin{aligned} \int\limits_E|f|\Delta\mu_\Delta=& \int\limits_E (|f|-g_N)\Delta\mu_\Delta+\int\limits_E g_N \Delta\mu_\Delta\\ \leq& \int\limits_{\mathbb{T}}(|f|-g_N)\Delta \mu_\Delta +\int\limits_E N \Delta\mu_\Delta\\ =& \int\limits_{\mathbb{T}}(|f|-g_N)\Delta\mu_\Delta +N\mu_\Delta(E)\\ <& \frac{\epsilon}{2}+N\mu_\Delta(E). \end{aligned} \notag\]
Now, we take \(\delta=\frac{\epsilon}{2N}>0\). Then, for any \(E\subset \mathbb{M}(m^*)\) with \(\mu_\Delta(E)<\delta\), we find
\[ \begin{aligned} \int\limits_E |f|\Delta\mu_\Delta<&\frac{\epsilon}{2}+N\left(\frac{\epsilon}{2N}\right)\\ =& \frac{\epsilon}{2}+\frac{\epsilon}{2}\\ =& \epsilon. \end{aligned} \notag\]
This completes the proof.
Let \(\{f_n\}_{n\in \mathbb{N}}\) be a sequence of Lebesgue delta measurable functions where \(f_n: \mathbb{T}\to \widetilde{\mathbb{R}}\) and \(f_n\to f\) on \(\mathbb{T}\) pointwise to some function \(f: \mathbb{T}\to \widetilde{\mathbb{R}}\). Suppose that there exists a function \(f\in \mathcal{L}^1_\Delta(\mathbb{T})\) such that \(|f_n|\leq g\) \(\mu_\Delta\)-a.e. on \(\mathbb{T}\) for any \(n\in \mathbb{N}\). Then \(f_n, f\in \mathcal{L}^1_\Delta(\mathbb{T})\) and
\[ \lim\limits_{n\to\infty} \int\limits_{\mathbb{T}}f_n\Delta\mu_\Delta= \int\limits_{\mathbb{T}}f\Delta\mu_\Delta. \notag\]
Зиддо
Since \(f_n\), \(n\in \mathbb{N}\), are Lebesgue delta measurable, we have that the limiting function \(f\) is Lebesgue delta measurable. By comparison, since \(|f_n|\leq g\) \(\mu_\Delta\)-a.e., \(f_n\), \(n\in \mathbb{N}\), are Lebesgue delta integrable over \(\mathbb{T}\) for any \(n\in \mathbb{N}\). Next,
\[ \begin{aligned} |f|=& \lim\limits_{n\to\infty}|f_n|\\ \leq& g\quad \mu_\Delta-\mbox{a.e.} \end{aligned} \notag\]
Hence, the function \(f\) must be Lebesgue delta integrable over \(\mathbb{T}\), as well. Define
\[ \begin{aligned} h_n=& g-f_n,\\ p_n=& g+f_n,\quad n\in \mathbb{N}. \end{aligned} \notag\]
We apply the Fatou lemma to the sequences \(\{h_n\}_{n\in \mathbb{N}}\}\) and \(\{p_n\}_{n\in \mathbb{N}}\) and we find
\[ \begin{aligned} \int\limits_{\mathbb{T}}(g-f)\Delta\mu_\Delta=& \int\limits_{\mathbb{T}}\liminf\limits_{n\to\infty} (g-f_n)\Delta\mu_\Delta\\ \leq& \liminf\limits_{n\to\infty} \int\limits_{\mathbb{T}}(g-f_n)\Delta\mu_\Delta\\ =& \int\limits_{\mathbb{T}}g\Delta\mu_\Delta -\limsup\limits_{n\to\infty}\int\limits_{\mathbb{T}}f_n\Delta\mu_\Delta, \end{aligned} \notag\]
and
\[ \begin{aligned} \int\limits_{\mathbb{T}}(g+f)\Delta\mu_\Delta=& \int\limits_{\mathbb{T}}\liminf\limits_{n\to\infty} (g+f_n)\Delta\mu_\Delta\\ \leq& \liminf\limits_{n\to\infty} \int\limits_{\mathbb{T}}(g+f_n)\Delta\mu_\Delta\\ =& \int\limits_{\mathbb{T}}g\Delta\mu_\Delta +\liminf\limits_{n\to\infty}\int\limits_{\mathbb{T}}f_n\Delta\mu_\Delta. \end{aligned} \notag\]
Therefore
\[ \begin{aligned} \limsup\limits_{n\to\infty} \int\limits_{\mathbb{T}}f_n\Delta\mu_\Delta\leq & \int\limits_{\mathbb{T}}f\Delta\mu_\Delta\\ \leq& \liminf\limits_{n\to\infty}\int\limits_{\mathbb{T}}f_n\Delta\mu_\Delta\\ \leq& \limsup\limits_{n\to\infty}\int\limits_{\mathbb{T}}f_n\Delta\mu_\Delta, \end{aligned} \notag\]
whereupon we get the desired result. \end{proof}