3

I would like to create this Homework Solutions and Assignment Sheets which was made it by adobe indesign

Could someone create it with latex ?

enter image description here

enter image description here

\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}
  • 3
    Your code contains errors which make it not compilable. Please, correct them. Once corrected (under my own criteria) it seems to produce something similar to the desired figure, long boxes followed by two one column boxes. If this is the desired result, what's the problem? – Ignasi Oct 2 '17 at 14:28
  • I update my code it's compilable now but still i cant create the same design as adobe indesign. could you please create it with latex – Educ Oct 3 '17 at 9:20
3

More or less looks like the desired format:

enter image description here

I've modified provided \centerbox to make it breakable and without title (original code included it, but it was not shown in final format). The rounded corners frame is added with overlay options. Separation between boxes is fixed with beforeafter skip=0pt.

Two columns boxes are created with a sidebyside box. It's not breakable. Left contents is inserted before \tcblower and right one after it. If box is larger than available space it will jump to next page.

\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}

Update:

The inclusion of boxed title makes me change overlay definition to avoid inner border of centerbox goes over the title. Now exercises use an auto counter which is shown inside title box. There's no space between exercise and solution but the default space between solution and next exercise.

enter image description here

\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}
  • Thank you could you please add the name exercise with counter – Educ Oct 3 '17 at 11:28
  • could you also make solution breakable – Educ Oct 3 '17 at 11:30
  • 1
    @Educ I suppose centerbox correspond to exercise. How do you want the title, like in original code? Do you want the double border (sharp+round)? – Ignasi Oct 3 '17 at 11:44
  • 1
    @Educ I don't know how to make solution breakable. Even being a raster it won't break boxes. – Ignasi Oct 3 '17 at 12:08
  • 1
    @Educ: I've updated the answer. – Ignasi Oct 3 '17 at 15:33

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