Indesign 的家庭作业解决方案和作业表

Question

我想创建这个由 Adob​​e Indesign 制作的家庭作业解决方案和作业表

有人可以用乳胶制作它吗？

\documentclass[11pt,a4paper]{report}
\usepackage[margin=0.5in]{geometry}
\usepackage{multicol}
\usepackage[french]{babel}
\usepackage{fontspec}
\usepackage{graphicx}
\usepackage{amsthm,amssymb,amsfonts,mathtools,lipsum,calc}
\usepackage{tcolorbox}
\usepackage{xhfill}
\tcbuselibrary{skins,raster}
\newtcolorbox{centerbox}[2][]{colback=white, colframe=black!75,fonttitle=\normalfont,
  colbacktitle=white, coltitle=black,
  enhanced, attach boxed title to top center={yshift=-3mm},
  title=#2,#1}

\begin{document}

\begin{centerbox}{Exercice 1}%
  Let $(g_k)\in\mathcal{R}^{*}(I)$ be a sequence that converges uniformly a.e. to g ( in the sense that there exists a null set $Z\subset I:[a,b]$ such that $(g_k)$ converges uniformly to g on $I\setminus Z$). Then $g \in\mathcal{R}^{*}(I)$ and (*) $\int_{I}g=\lim\int_{I} g_k$
  \end{centerbox}

\begin{tcbraster}[raster columns=2,raster equal height=rows,raster valign=top,raster column skip=5mm]

Let $f_{k}(x):=g_{k}(x)$ and $f(x):=g(x)$ for $x\in I\setminus Z$ and $:=0$ on $Z$. By Theorem 8.3 and Exercise 3.C we have $\int_{I}g_k=\int_{I}f_k=\int_{k}f=\int_{I}g$. 

Let $f_{k}(x):=g_{k}(x)$ and $f(x):=g(x)$ for $x\in I\setminus Z$ and $:=0$ on $Z$. By Theorem 8.3 and Exercise 3.C we have $\int_{I}g_k=\int_{I}f_k=\int_{k}f=\int_{I}g$.


Let $f_{k}(x):=g_{k}(x)$ and $f(x):=g(x)$ for $x\in I\setminus Z$ and $:=0$ on $Z$. By Theorem 8.3 and Exercise 3.C we have $\int_{I}g_k=\int_{I}f_k=\int_{k}f=\int_{I}g$.


Let $f_{k}(x):=g_{k}(x)$ and $f(x):=g(x)$ for $x\in I\setminus Z$ and $:=0$ on $Z$. By Theorem 8.3 and Exercise 3.C we have $\int_{I}g_k=\int_{I}f_k=\int_{k}f=\int_{I}g$. 
    \end{tcbraster}

\begin{centerbox}{Exercice 1}%
  Let $(g_k)\in\mathcal{R}^{*}(I)$ be a sequence that converges uniformly a.e. to g ( in the sense that there exists a null set $Z\subset I:[a,b]$ such that $(g_k)$ converges uniformly to g on $I\setminus Z$). Then $g \in\mathcal{R}^{*}(I)$ and (*) $\int_{I}g=\lim\int_{I} g_k$
  \end{centerbox}

\begin{tcbraster}[raster columns=2,raster equal height=rows,raster valign=top,raster column skip=5mm]

Let $f_{k}(x):=g_{k}(x)$ and $f(x):=g(x)$ for $x\in I\setminus Z$ and $:=0$ on $Z$. By Theorem 8.3 and Exercise 3.C we have $\int_{I}g_k=\int_{I}f_k=\int_{k}f=\int_{I}g$. 

Let $f_{k}(x):=g_{k}(x)$ and $f(x):=g(x)$ for $x\in I\setminus Z$ and $:=0$ on $Z$. By Theorem 8.3 and Exercise 3.C we have $\int_{I}g_k=\int_{I}f_k=\int_{k}f=\int_{I}g$.


Let $f_{k}(x):=g_{k}(x)$ and $f(x):=g(x)$ for $x\in I\setminus Z$ and $:=0$ on $Z$. By Theorem 8.3 and Exercise 3.C we have $\int_{I}g_k=\int_{I}f_k=\int_{k}f=\int_{I}g$.


Let $f_{k}(x):=g_{k}(x)$ and $f(x):=g(x)$ for $x\in I\setminus Z$ and $:=0$ on $Z$. By Theorem 8.3 and Exercise 3.C we have $\int_{I}g_k=\int_{I}f_k=\int_{k}f=\int_{I}g$. 
    \end{tcbraster}

\begin{centerbox}{Exercice 1}%
  Let $(g_k)\in\mathcal{R}^{*}(I)$ be a sequence that converges uniformly a.e. to g ( in the sense that there exists a null set $Z\subset I:[a,b]$ such that $(g_k)$ converges uniformly to g on $I\setminus Z$). Then $g \in\mathcal{R}^{*}(I)$ and (*) $\int_{I}g=\lim\int_{I} g_k$
  \end{centerbox}

\begin{tcbraster}[raster columns=2,raster equal height=rows,raster valign=top,raster column skip=5mm]

Let $f_{k}(x):=g_{k}(x)$ and $f(x):=g(x)$ for $x\in I\setminus Z$ and $:=0$ on $Z$. By Theorem 8.3 and Exercise 3.C we have $\int_{I}g_k=\int_{I}f_k=\int_{k}f=\int_{I}g$. 

Let $f_{k}(x):=g_{k}(x)$ and $f(x):=g(x)$ for $x\in I\setminus Z$ and $:=0$ on $Z$. By Theorem 8.3 and Exercise 3.C we have $\int_{I}g_k=\int_{I}f_k=\int_{k}f=\int_{I}g$.


Let $f_{k}(x):=g_{k}(x)$ and $f(x):=g(x)$ for $x\in I\setminus Z$ and $:=0$ on $Z$. By Theorem 8.3 and Exercise 3.C we have $\int_{I}g_k=\int_{I}f_k=\int_{k}f=\int_{I}g$.


Let $f_{k}(x):=g_{k}(x)$ and $f(x):=g(x)$ for $x\in I\setminus Z$ and $:=0$ on $Z$. By Theorem 8.3 and Exercise 3.C we have $\int_{I}g_k=\int_{I}f_k=\int_{k}f=\int_{I}g$. 
    \end{tcbraster}

\end{document}

Answer 1

或多或少看起来像所需的格式：

我已经修改了提供的，\centerbox使其变为有的breakable和没有的title（原始代码包括它，但它没有以最终格式显示）。框架rounded corners添加了overlay选项。框之间的分隔用固定beforeafter skip=0pt。

用一个框创建两列框sidebyside。它不可破坏。左侧内容插入在它之前\tcblower，右侧内容插入在它之后。如果框大于可用空间，它将跳转到下一页。

\documentclass[11pt,a4paper]{report}
\usepackage[margin=0.5in]{geometry}
\usepackage{multicol}
\usepackage[french]{babel}
\usepackage{fontspec}
\usepackage{graphicx}
\usepackage{amsthm,amssymb,amsfonts,mathtools,lipsum,calc}
\usepackage[most]{tcolorbox}
\usepackage{xhfill}
%\tcbuselibrary{skins,raster}
\tcbset{beforeafter skip=0pt}
\newtcolorbox{centerbox}[1][]{colback=white, colframe=black!75, 
    % fonttitle=\normalfont,
   % colbacktitle=white, coltitle=black,
  enhanced, breakable,
    % attach boxed title to top center={yshift=-3mm},
    notitle,
    %enlarge by=1mm, 
    sharp corners,
    boxsep=3mm,
%   show bounding box,
    overlay unbroken={\draw[rounded corners, line width=.5mm] ([shift={(2mm,2mm)}]frame.south west) rectangle ([shift={(-2mm,-2mm)}]frame.north east);},
    overlay first={\draw[rounded corners, line width=.5mm] ([xshift=2mm]frame.south west) |- ([yshift=-2mm]frame.north) -| ([xshift=-2mm]frame.south east);},
    overlay middle={\draw[rounded corners, line width=.5mm] ([xshift=2mm]frame.south west)--([xshift=2mm]frame.north west);
    \draw[rounded corners, line width=.5mm] ([xshift=-2mm]frame.south east)--([xshift=-2mm]frame.north east);},
    overlay last={\draw[rounded corners, line width=.5mm] ([xshift=2mm]frame.north west) |- ([yshift=2mm]frame.south) -| ([xshift=-2mm]frame.north east);},
%   borderline={.5pt}{1mm}{blue, rounded corners}
    %title=#2,
    #1}
\newtcolorbox{twoside}[1][]{colback=white, colframe=black!75, fonttitle=\normalfont,
  sharp corners, sidebyside, 
  enhanced, notitle, segmentation style={double=white, draw=black!75, solid, line width=.5mm},#1}

\begin{document}

\begin{centerbox}%
  Let $(g_k)\in\mathcal{R}^{*}(I)$ be a sequence that converges uniformly a.e. to g ( in the sense that there exists a null set $Z\subset I:[a,b]$ such that $(g_k)$ converges uniformly to g on $I\setminus Z$). Then $g \in\mathcal{R}^{*}(I)$ and (*) $\int_{I}g=\lim\int_{I} g_k$
  \end{centerbox}

\begin{twoside}
Let $f_{k}(x):=g_{k}(x)$ and $f(x):=g(x)$ for $x\in I\setminus Z$ and $:=0$ on $Z$. By Theorem 8.3 and Exercise 3.C we have $\int_{I}g_k=\int_{I}f_k=\int_{k}f=\int_{I}g$. 

Let $f_{k}(x):=g_{k}(x)$ and $f(x):=g(x)$ for $x\in I\setminus Z$ and $:=0$ on $Z$. By Theorem 8.3 and Exercise 3.C we have $\int_{I}g_k=\int_{I}f_k=\int_{k}f=\int_{I}g$.

\tcblower

Let $f_{k}(x):=g_{k}(x)$ and $f(x):=g(x)$ for $x\in I\setminus Z$ and $:=0$ on $Z$. By Theorem 8.3 and Exercise 3.C we have $\int_{I}g_k=\int_{I}f_k=\int_{k}f=\int_{I}g$.


Let $f_{k}(x):=g_{k}(x)$ and $f(x):=g(x)$ for $x\in I\setminus Z$ and $:=0$ on $Z$. By Theorem 8.3 and Exercise 3.C we have $\int_{I}g_k=\int_{I}f_k=\int_{k}f=\int_{I}g$. 
\end{twoside}

\begin{centerbox}%
  Let $(g_k)\in\mathcal{R}^{*}(I)$ be a sequence that converges uniformly a.e. to g ( in the sense that there exists a null set $Z\subset I:[a,b]$ such that $(g_k)$ converges uniformly to g on $I\setminus Z$). Then $g \in\mathcal{R}^{*}(I)$ and (*) $\int_{I}g=\lim\int_{I} g_k$
  \end{centerbox}

\begin{twoside}
Let $f_{k}(x):=g_{k}(x)$ and $f(x):=g(x)$ for $x\in I\setminus Z$ and $:=0$ on $Z$. By Theorem 8.3 and Exercise 3.C we have $\int_{I}g_k=\int_{I}f_k=\int_{k}f=\int_{I}g$. 

Let $f_{k}(x):=g_{k}(x)$ and $f(x):=g(x)$ for $x\in I\setminus Z$ and $:=0$ on $Z$. By Theorem 8.3 and Exercise 3.C we have $\int_{I}g_k=\int_{I}f_k=\int_{k}f=\int_{I}g$.

\tcblower

Let $f_{k}(x):=g_{k}(x)$ and $f(x):=g(x)$ for $x\in I\setminus Z$ and $:=0$ on $Z$. By Theorem 8.3 and Exercise 3.C we have $\int_{I}g_k=\int_{I}f_k=\int_{k}f=\int_{I}g$.


Let $f_{k}(x):=g_{k}(x)$ and $f(x):=g(x)$ for $x\in I\setminus Z$ and $:=0$ on $Z$. By Theorem 8.3 and Exercise 3.C we have $\int_{I}g_k=\int_{I}f_k=\int_{k}f=\int_{I}g$. 
\end{twoside}

\begin{centerbox}%
  Let $(g_k)\in\mathcal{R}^{*}(I)$ be a sequence that converges uniformly a.e. to g ( in the sense that there exists a null set $Z\subset I:[a,b]$ such that $(g_k)$ converges uniformly to g on $I\setminus Z$). Then $g \in\mathcal{R}^{*}(I)$ and (*) $\int_{I}g=\lim\int_{I} g_k$
  \end{centerbox}

\begin{twoside}
Let $f_{k}(x):=g_{k}(x)$ and $f(x):=g(x)$ for $x\in I\setminus Z$ and $:=0$ on $Z$. By Theorem 8.3 and Exercise 3.C we have $\int_{I}g_k=\int_{I}f_k=\int_{k}f=\int_{I}g$. 

Let $f_{k}(x):=g_{k}(x)$ and $f(x):=g(x)$ for $x\in I\setminus Z$ and $:=0$ on $Z$. By Theorem 8.3 and Exercise 3.C we have $\int_{I}g_k=\int_{I}f_k=\int_{k}f=\int_{I}g$.

\tcblower

Let $f_{k}(x):=g_{k}(x)$ and $f(x):=g(x)$ for $x\in I\setminus Z$ and $:=0$ on $Z$. By Theorem 8.3 and Exercise 3.C we have $\int_{I}g_k=\int_{I}f_k=\int_{k}f=\int_{I}g$.


Let $f_{k}(x):=g_{k}(x)$ and $f(x):=g(x)$ for $x\in I\setminus Z$ and $:=0$ on $Z$. By Theorem 8.3 and Exercise 3.C we have $\int_{I}g_k=\int_{I}f_k=\int_{k}f=\int_{I}g$. 
\end{twoside}

\begin{centerbox}%
  Let $(g_k)\in\mathcal{R}^{*}(I)$ be a sequence that converges uniformly a.e. to g ( in the sense that there exists a null set $Z\subset I:[a,b]$ such that $(g_k)$ converges uniformly to g on $I\setminus Z$). Then $g \in\mathcal{R}^{*}(I)$ and (*) $\int_{I}g=\lim\int_{I} g_k$

  Let $(g_k)\in\mathcal{R}^{*}(I)$ be a sequence that converges uniformly a.e. to g ( in the sense that there exists a null set $Z\subset I:[a,b]$ such that $(g_k)$ converges uniformly to g on $I\setminus Z$). Then $g \in\mathcal{R}^{*}(I)$ and (*) $\int_{I}g=\lim\int_{I} g_k$
  Let $(g_k)\in\mathcal{R}^{*}(I)$ be a sequence that converges uniformly a.e. to g ( in the sense that there exists a null set $Z\subset I:[a,b]$ such that $(g_k)$ converges uniformly to g on $I\setminus Z$). Then $g \in\mathcal{R}^{*}(I)$ and (*) $\int_{I}g=\lim\int_{I} g_k$

  Let $(g_k)\in\mathcal{R}^{*}(I)$ be a sequence that converges uniformly a.e. to g ( in the sense that there exists a null set $Z\subset I:[a,b]$ such that $(g_k)$ converges uniformly to g on $I\setminus Z$). Then $g \in\mathcal{R}^{*}(I)$ and (*) $\int_{I}g=\lim\int_{I} g_k$

  Let $(g_k)\in\mathcal{R}^{*}(I)$ be a sequence that converges uniformly a.e. to g ( in the sense that there exists a null set $Z\subset I:[a,b]$ such that $(g_k)$ converges uniformly to g on $I\setminus Z$). Then $g \in\mathcal{R}^{*}(I)$ and (*) $\int_{I}g=\lim\int_{I} g_k$

  Let $(g_k)\in\mathcal{R}^{*}(I)$ be a sequence that converges uniformly a.e. to g ( in the sense that there exists a null set $Z\subset I:[a,b]$ such that $(g_k)$ converges uniformly to g on $I\setminus Z$). Then $g \in\mathcal{R}^{*}(I)$ and (*) $\int_{I}g=\lim\int_{I} g_k$

  Let $(g_k)\in\mathcal{R}^{*}(I)$ be a sequence that converges uniformly a.e. to g ( in the sense that there exists a null set $Z\subset I:[a,b]$ such that $(g_k)$ converges uniformly to g on $I\setminus Z$). Then $g \in\mathcal{R}^{*}(I)$ and (*) $\int_{I}g=\lim\int_{I} g_k$

  Let $(g_k)\in\mathcal{R}^{*}(I)$ be a sequence that converges uniformly a.e. to g ( in the sense that there exists a null set $Z\subset I:[a,b]$ such that $(g_k)$ converges uniformly to g on $I\setminus Z$). Then $g \in\mathcal{R}^{*}(I)$ and (*) $\int_{I}g=\lim\int_{I} g_k$

  Let $(g_k)\in\mathcal{R}^{*}(I)$ be a sequence that converges uniformly a.e. to g ( in the sense that there exists a null set $Z\subset I:[a,b]$ such that $(g_k)$ converges uniformly to g on $I\setminus Z$). Then $g \in\mathcal{R}^{*}(I)$ and (*) $\int_{I}g=\lim\int_{I} g_k$

  Let $(g_k)\in\mathcal{R}^{*}(I)$ be a sequence that converges uniformly a.e. to g ( in the sense that there exists a null set $Z\subset I:[a,b]$ such that $(g_k)$ converges uniformly to g on $I\setminus Z$). Then $g \in\mathcal{R}^{*}(I)$ and (*) $\int_{I}g=\lim\int_{I} g_k$

  Let $(g_k)\in\mathcal{R}^{*}(I)$ be a sequence that converges uniformly a.e. to g ( in the sense that there exists a null set $Z\subset I:[a,b]$ such that $(g_k)$ converges uniformly to g on $I\setminus Z$). Then $g \in\mathcal{R}^{*}(I)$ and (*) $\int_{I}g=\lim\int_{I} g_k$

  Let $(g_k)\in\mathcal{R}^{*}(I)$ be a sequence that converges uniformly a.e. to g ( in the sense that there exists a null set $Z\subset I:[a,b]$ such that $(g_k)$ converges uniformly to g on $I\setminus Z$). Then $g \in\mathcal{R}^{*}(I)$ and (*) $\int_{I}g=\lim\int_{I} g_k$

  Let $(g_k)\in\mathcal{R}^{*}(I)$ be a sequence that converges uniformly a.e. to g ( in the sense that there exists a null set $Z\subset I:[a,b]$ such that $(g_k)$ converges uniformly to g on $I\setminus Z$). Then $g \in\mathcal{R}^{*}(I)$ and (*) $\int_{I}g=\lim\int_{I} g_k$

  Let $(g_k)\in\mathcal{R}^{*}(I)$ be a sequence that converges uniformly a.e. to g ( in the sense that there exists a null set $Z\subset I:[a,b]$ such that $(g_k)$ converges uniformly to g on $I\setminus Z$). Then $g \in\mathcal{R}^{*}(I)$ and (*) $\int_{I}g=\lim\int_{I} g_k$

  Let $(g_k)\in\mathcal{R}^{*}(I)$ be a sequence that converges uniformly a.e. to g ( in the sense that there exists a null set $Z\subset I:[a,b]$ such that $(g_k)$ converges uniformly to g on $I\setminus Z$). Then $g \in\mathcal{R}^{*}(I)$ and (*) $\int_{I}g=\lim\int_{I} g_k$
  \end{centerbox}

\end{document}

更新：

包含框状标题让我改变了overlay定义，以避免内边框centerbox超出标题。现在练习使用auto counter显示在标题框内的。练习和解决方案之间没有空格，但解决方案和下一个练习之间有默认空格。

\documentclass[11pt,a4paper]{report}
\usepackage[margin=0.5in]{geometry}
\usepackage{multicol}
\usepackage[french]{babel}
\usepackage{fontspec}
\usepackage{graphicx}
\usepackage{amsthm,amssymb,amsfonts,mathtools,lipsum,calc}
\usepackage[most]{tcolorbox}
\usepackage{xhfill}
\usepackage{multicol}

%\tcbuselibrary{skins,raster}
%\tcbset{beforeafter skip=0pt}
\newtcolorbox[auto counter]{centerbox}[1][]{
    colback=white, 
    colframe=black!75, 
    fonttitle=\normalfont,
   colbacktitle=white, coltitle=black,
   enhanced, breakable,
    attach boxed title to top center={yshift=-\tcboxedtitleheight/2},
    title=Exercise \thetcbcounter,
    sharp corners,
    boxsep=3mm,
    after skip=0pt,
    underlay unbroken={\draw[rounded corners, line width=.5mm] ([yshift=-2mm]title.west) -| ([shift={(2mm,2mm)}]frame.south west)-|([shift={(-2mm,-2mm)}]frame.north east)--([yshift=-2mm]title.east);},
    overlay first={\draw[rounded corners, line width=.5mm] ([xshift=2mm]frame.south west) |- ([yshift=-2mm]title.west);
    \draw[rounded corners, line width=.5mm] ([yshift=-2mm]title.east) -| ([xshift=-2mm]frame.south east);},
    overlay middle={\draw[rounded corners, line width=.5mm] ([xshift=2mm]frame.south west)--([xshift=2mm]frame.north west);
    \draw[rounded corners, line width=.5mm] ([xshift=-2mm]frame.south east)--([xshift=-2mm]frame.north east);},
    overlay last={\draw[rounded corners, line width=.5mm] ([xshift=2mm]frame.north west) |- ([yshift=2mm]frame.south) -| ([xshift=-2mm]frame.north east);},
    #1}
\newtcolorbox{twoside}[1][]{colback=white, colframe=black!75, fonttitle=\normalfont,
  sharp corners, sidebyside, before skip=-.5mm,
  enhanced, notitle, segmentation style={double=white, draw=black!75, solid, line width=.5mm},#1}

\begin{document}

\begin{centerbox}%
  Let $(g_k)\in\mathcal{R}^{*}(I)$ be a sequence that converges uniformly a.e. to g ( in the sense that there exists a null set $Z\subset I:[a,b]$ such that $(g_k)$ converges uniformly to g on $I\setminus Z$). Then $g \in\mathcal{R}^{*}(I)$ and (*) $\int_{I}g=\lim\int_{I} g_k$
  \end{centerbox}

\begin{twoside}
Let $f_{k}(x):=g_{k}(x)$ and $f(x):=g(x)$ for $x\in I\setminus Z$ and $:=0$ on $Z$. By Theorem 8.3 and Exercise 3.C we have $\int_{I}g_k=\int_{I}f_k=\int_{k}f=\int_{I}g$. 

Let $f_{k}(x):=g_{k}(x)$ and $f(x):=g(x)$ for $x\in I\setminus Z$ and $:=0$ on $Z$. By Theorem 8.3 and Exercise 3.C we have $\int_{I}g_k=\int_{I}f_k=\int_{k}f=\int_{I}g$.

\tcblower

Let $f_{k}(x):=g_{k}(x)$ and $f(x):=g(x)$ for $x\in I\setminus Z$ and $:=0$ on $Z$. By Theorem 8.3 and Exercise 3.C we have $\int_{I}g_k=\int_{I}f_k=\int_{k}f=\int_{I}g$.


Let $f_{k}(x):=g_{k}(x)$ and $f(x):=g(x)$ for $x\in I\setminus Z$ and $:=0$ on $Z$. By Theorem 8.3 and Exercise 3.C we have $\int_{I}g_k=\int_{I}f_k=\int_{k}f=\int_{I}g$. 
\end{twoside}

\begin{centerbox}%
  Let $(g_k)\in\mathcal{R}^{*}(I)$ be a sequence that converges uniformly a.e. to g ( in the sense that there exists a null set $Z\subset I:[a,b]$ such that $(g_k)$ converges uniformly to g on $I\setminus Z$). Then $g \in\mathcal{R}^{*}(I)$ and (*) $\int_{I}g=\lim\int_{I} g_k$
  \end{centerbox}

\begin{twoside}
Let $f_{k}(x):=g_{k}(x)$ and $f(x):=g(x)$ for $x\in I\setminus Z$ and $:=0$ on $Z$. By Theorem 8.3 and Exercise 3.C we have $\int_{I}g_k=\int_{I}f_k=\int_{k}f=\int_{I}g$. 

Let $f_{k}(x):=g_{k}(x)$ and $f(x):=g(x)$ for $x\in I\setminus Z$ and $:=0$ on $Z$. By Theorem 8.3 and Exercise 3.C we have $\int_{I}g_k=\int_{I}f_k=\int_{k}f=\int_{I}g$.

\tcblower

Let $f_{k}(x):=g_{k}(x)$ and $f(x):=g(x)$ for $x\in I\setminus Z$ and $:=0$ on $Z$. By Theorem 8.3 and Exercise 3.C we have $\int_{I}g_k=\int_{I}f_k=\int_{k}f=\int_{I}g$.


Let $f_{k}(x):=g_{k}(x)$ and $f(x):=g(x)$ for $x\in I\setminus Z$ and $:=0$ on $Z$. By Theorem 8.3 and Exercise 3.C we have $\int_{I}g_k=\int_{I}f_k=\int_{k}f=\int_{I}g$. 
\end{twoside}

\begin{centerbox}%
  Let $(g_k)\in\mathcal{R}^{*}(I)$ be a sequence that converges uniformly a.e. to g ( in the sense that there exists a null set $Z\subset I:[a,b]$ such that $(g_k)$ converges uniformly to g on $I\setminus Z$). Then $g \in\mathcal{R}^{*}(I)$ and (*) $\int_{I}g=\lim\int_{I} g_k$
  \end{centerbox}

\begin{twoside}
Let $f_{k}(x):=g_{k}(x)$ and $f(x):=g(x)$ for $x\in I\setminus Z$ and $:=0$ on $Z$. By Theorem 8.3 and Exercise 3.C we have $\int_{I}g_k=\int_{I}f_k=\int_{k}f=\int_{I}g$. 

Let $f_{k}(x):=g_{k}(x)$ and $f(x):=g(x)$ for $x\in I\setminus Z$ and $:=0$ on $Z$. By Theorem 8.3 and Exercise 3.C we have $\int_{I}g_k=\int_{I}f_k=\int_{k}f=\int_{I}g$.

\tcblower

Let $f_{k}(x):=g_{k}(x)$ and $f(x):=g(x)$ for $x\in I\setminus Z$ and $:=0$ on $Z$. By Theorem 8.3 and Exercise 3.C we have $\int_{I}g_k=\int_{I}f_k=\int_{k}f=\int_{I}g$.


Let $f_{k}(x):=g_{k}(x)$ and $f(x):=g(x)$ for $x\in I\setminus Z$ and $:=0$ on $Z$. By Theorem 8.3 and Exercise 3.C we have $\int_{I}g_k=\int_{I}f_k=\int_{k}f=\int_{I}g$. 
\end{twoside}

\end{document}