Chapter 9: The Lebesgue Delta Integral
- Page ID
- 208229
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)The construction of the Lebesgue delta integral can be generalised from nonnegative functions to any Lebesgue delta measurable functions with image in \(\widetilde{\mathbb{R}}=[-\infty, \infty]\). This is done by spliting an arbitrary Lebesgue delta measurable function \(f: \mathbb{T}\to \widetilde{\mathbb{R}}\) into its positive and negative parts, namely
\[ f=f^+-f^-, \notag\]
which are two nonnegative functions.
If at least one of the two integrals
\[ \int\limits_{\mathbb{T}}f^+\Delta\mu_\Delta\quad \mbox{and}\quad \int\limits_{\mathbb{T}}f^-\Delta \mu_\Delta \notag\]
is finite, we can define the Lebesgue delta integral as follows
\[ \int\limits_{\mathbb{T}}f\Delta\mu_\Delta=\int\limits_{\mathbb{T}}f^+\Delta\mu_\Delta-\int\limits_{\mathbb{T}}f^-\Delta \mu_\Delta. \notag\]
If both integrals are finite, then we say that the function \(f\) is Lebesgue delta integrable over \(\mathbb{T}\). The space of all Lebesgue delta integrable functions over \(\mathbb{T}\) is denoted by \(\mathcal{L}^1_\Delta(\mathbb{T})\).
Suppose that \(f, g:\mathbb{T}\to \widetilde{\mathbb{R}}\) are Lebesgue delta measurable functions such that \(f, g\in \mathcal{L}_\Delta^1(\mathbb{T})\) and \(\alpha, \beta\in \mathbb{R}\) are constants. Then, we have the following.
1. \(\alpha f\in \mathcal{L}_\Delta^1(\mathbb{T})\) and
\[ \int\limits_{\mathbb{T}}\alpha f\Delta\mu_\Delta= \alpha \int\limits_{\mathbb{T}}f\Delta \mu_\Delta. \notag\]
2. \(\alpha f+\beta g\in \mathcal{L}^1_\Delta(\mathbb{T})\) and
\[ \int\limits_{\mathbb{T}}(\alpha f+\beta g)\Delta\mu_\Delta=\alpha\int\limits_{\mathbb{T}} f\Delta\mu_\Delta+\beta \int\limits_{\mathbb{T}}g \Delta\mu_\Delta. \notag\]
Proof
1. If \(\alpha=0\), then there is nothing to prove. Otherwise, we have two cases.
(a) \(\alpha>0\). Then \
\[ \begin{aligned} (\alpha f)^+=& \alpha f^+,\\ (\alpha f)^-=& \alpha f^-. \end{aligned} \notag\]
Hence, using the homogeneity of the Lebesgue delta integral of nonnegative Lebesgue delta measurable functions, we arrive at
\[ \begin{aligned} \int\limits_{\mathbb{T}}\alpha f\Delta\mu_\Delta=& \int\limits_{\mathbb{T}}(\alpha f)^+ \Delta\mu_\Delta-\int\limits_{\mathbb{T}}(\alpha f)^-\Delta \mu_\Delta\\ =& \int\limits_{\mathbb{T}}\alpha f^+\Delta \mu_\Delta -\int\limits_{\mathbb{T}}\alpha f^-\Delta\mu_\Delta\\ =& \alpha\int\limits_{\mathbb{T}} f^+\Delta\mu_\Delta-\alpha \int\limits_{\mathbb{T}}f^-\Delta \mu_\Delta\\ =& \alpha\left(\int\limits_{\mathbb{T}}f^+\Delta\mu_\Delta -\int\limits_{\mathbb{T}}f^-\Delta \mu_\Delta\right)\\ =& \alpha \int\limits_{\mathbb{T}}f\Delta \mu_\Delta. \end{aligned} \notag\]
(b) \(\alpha<0\). Then
\[ \begin{aligned} (\alpha f)^+=& \max(\alpha f, 0)\\ =& \max((-\alpha)(-f), 0)\\ =& (-\alpha)\max(-f, 0)\\ =& (-\alpha)f^-, \end{aligned} \notag\]
and likewise,
\[ (\alpha f)^-=(-\alpha) f^+. \notag\]
Hence, using that \(-\alpha>0\), we get
\[ \begin{aligned} \int\limits_{\mathbb{T}}\alpha f\Delta\mu_\Delta=& \int\limits_{\mathbb{T}}(\alpha f)^+ \Delta\mu_\Delta -\int\limits_{\mathbb{T}}(\alpha f)^- \Delta \mu_\Delta\\ =& \int\limits_{\mathbb{T}}(-\alpha)f^-\Delta\mu_\Delta-\int\limits_{\mathbb{T}}(-\alpha)f^+\Delta\mu_\Delta\\ =& (-\alpha)\int\limits_{\mathbb{T}}f^-\Delta\mu_\Delta+\alpha\int\limits_{\mathbb{T}}f^+\Delta\mu_\Delta\\ =& \alpha\left(\int\limits_{\mathbb{T}}f^+\Delta \mu_\Delta -\int\limits_{\mathbb{T}}f^-\Delta\mu_\Delta\right)\\ =& \alpha \int\limits_{\mathbb{T}}f\Delta\mu_\Delta. \end{aligned} \notag\]
2. First, assume that \(\alpha=\beta=1\). We have
\[ f+g=(f+g)^+-(f+g)^- \notag\]
and
\[ f+g=(f^+-f^-)+(g^+-g^-). \notag\]
Combining this, we find
\[ (f^+-f^-)+(g^+-g^-)=(f+g)^+-(f+g)^-, \notag\]
or
\[ f^++g^++(f+g)^-=(f+g)^++f^-+g^-, \notag\]
where all terms of both sides are nonnegative.
Thus,
\[ \int\limits_{\mathbb{T}}\left((f+g)^++f^-+g^-\right)\Delta\mu_\Delta=\int\limits_{\mathbb{T}}\left(f^++g^++(f+g)^-\right)\Delta\mu_\Delta, \notag\]
or
\[ \int\limits_{\mathbb{T}}(f+g)^+\Delta\mu_\Delta+\int\limits_{\mathbb{T}}f^-\Delta\mu_\Delta +\int\limits_{\mathbb{T}}g^-\Delta\mu_\Delta= \int\limits_{\mathbb{T}}f^+\Delta\mu_\Delta +\int\limits_{\mathbb{T}}g^+\Delta \mu_\Delta +\int\limits_{\mathbb{T}}(f+g)^-\Delta\mu_\Delta, \notag\]
whereupon
\[ \int\limits_{\mathbb{T}}(f+g)^+\Delta\mu_\Delta -\int\limits_{\mathbb{T}}(f+g)^-\Delta \mu_\Delta =\left(\int\limits_{\mathbb{T}}f^+\Delta\mu_\Delta -\int\limits_{\mathbb{T}}f^-\Delta\mu_\Delta\right)+\left(\int\limits_{\mathbb{T}}g^+\Delta\mu_\Delta -\int\limits_{\mathbb{T}}g^-\Delta\mu_\Delta\right), \notag\]
or
\[ \int\limits_{\mathbb{T}}(f+g)\Delta\mu_\Delta=\int\limits_{\mathbb{T}}f\Delta\mu_\Delta +\int\limits_{\mathbb{T}}g\Delta \mu_\Delta. \notag\]
Now, we apply this for the functions \((\alpha f)\) and \((\beta g)\), and using the first assertion, we obtain
\[ \begin{aligned} \int\limits_{\mathbb{T}}((\alpha f)+(\beta g))\Delta\mu_\Delta=& \int\limits_{\mathbb{T}}\alpha f\Delta\mu_\Delta +\int\limits_{\mathbb{T}} \beta g\Delta \mu_\Delta\\ =& \alpha\int\limits_{\mathbb{T}}f\Delta\mu_\Delta+\beta \int\limits_{\mathbb{T}} g\Delta \mu_\Delta. \end{aligned} \notag\]
This completes the proof.