Prove that $f in R(F_s)$ on the interval $[a, b]$ and that $int f dF_s = f(s)$












0












$begingroup$


Let $a<s<b$ and let $f:[a, b] to mathbb{R}$ be a bounded function that is continous at the point s. Define



$F_s(x) = begin{cases}
0, & text{if $a le x lt s$} \
1, & text{if $s le x le b$}
end{cases}$



Prove that $f in R(F_s)$ on the interval $[a, b]$ and that



$int fdF_s = f(s)$.



My attempt to the proof:



Consider the partition,$P$ of the interval $[a, b]$ where $a=x_o le x_1 le ... le x_{i-1} le x_i le ... le x_n = b$.



Let $M_i = supf(x) and m_i = inf f(x)$ for $x_{i-1} le x le x_i$.



Given $epsilon gt 0$, and $P$. $fin R(F_s)$ if $U(P,f, F_s)-L(P,f,F_s) lt epsilon$.



We know $F(x)$ is continuous at $s in [a,b]$. Let there exist subinterval $[x_{i-1}, x_i]$ in $[a, b]$ such that $s in (x_{i-1}, x_i)$.



Let $F(x_i) - F(x_{i-1}) lt frac{epsilon}{2mM}$. Let $M_i - m_i lt 2M$. Then $sum_{i=0}^{n}(M_i - m_i) lt 2mM$ because there are exactly $m$ terms in this sum, as there are exactly $m$ of the intervals in $[x_{i-1}, x_i]$.



$U(P,f, F_s) - L(P,f, F_s) = (F(x_i)-F(x_{i-1}))sum_{i=0}^{n}(M_i - m_i) lt 2mM frac{epsilon}{2mM} = epsilon$.



Therefore, $f in R(F_s)$.



For the second part of the proof, we need to show: $int fdF_s = f(s)$.



Am I doing this proof right so far? could someone help me with the rest?



Thanks in advance.






proof-verification metric-spaces riemann-integration







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    Let $a<s<b$ and let $f:[a, b] to mathbb{R}$ be a bounded function that is continous at the point s. Define



    $F_s(x) = begin{cases}
    0, & text{if $a le x lt s$} \
    1, & text{if $s le x le b$}
    end{cases}$



    Prove that $f in R(F_s)$ on the interval $[a, b]$ and that



    $int fdF_s = f(s)$.



    My attempt to the proof:



    Consider the partition,$P$ of the interval $[a, b]$ where $a=x_o le x_1 le ... le x_{i-1} le x_i le ... le x_n = b$.



    Let $M_i = supf(x) and m_i = inf f(x)$ for $x_{i-1} le x le x_i$.



    Given $epsilon gt 0$, and $P$. $fin R(F_s)$ if $U(P,f, F_s)-L(P,f,F_s) lt epsilon$.



    We know $F(x)$ is continuous at $s in [a,b]$. Let there exist subinterval $[x_{i-1}, x_i]$ in $[a, b]$ such that $s in (x_{i-1}, x_i)$.



    Let $F(x_i) - F(x_{i-1}) lt frac{epsilon}{2mM}$. Let $M_i - m_i lt 2M$. Then $sum_{i=0}^{n}(M_i - m_i) lt 2mM$ because there are exactly $m$ terms in this sum, as there are exactly $m$ of the intervals in $[x_{i-1}, x_i]$.



    $U(P,f, F_s) - L(P,f, F_s) = (F(x_i)-F(x_{i-1}))sum_{i=0}^{n}(M_i - m_i) lt 2mM frac{epsilon}{2mM} = epsilon$.



    Therefore, $f in R(F_s)$.



    For the second part of the proof, we need to show: $int fdF_s = f(s)$.



    Am I doing this proof right so far? could someone help me with the rest?



    Thanks in advance.






    proof-verification metric-spaces riemann-integration







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Let $a<s<b$ and let $f:[a, b] to mathbb{R}$ be a bounded function that is continous at the point s. Define



      $F_s(x) = begin{cases}
      0, & text{if $a le x lt s$} \
      1, & text{if $s le x le b$}
      end{cases}$



      Prove that $f in R(F_s)$ on the interval $[a, b]$ and that



      $int fdF_s = f(s)$.



      My attempt to the proof:



      Consider the partition,$P$ of the interval $[a, b]$ where $a=x_o le x_1 le ... le x_{i-1} le x_i le ... le x_n = b$.



      Let $M_i = supf(x) and m_i = inf f(x)$ for $x_{i-1} le x le x_i$.



      Given $epsilon gt 0$, and $P$. $fin R(F_s)$ if $U(P,f, F_s)-L(P,f,F_s) lt epsilon$.



      We know $F(x)$ is continuous at $s in [a,b]$. Let there exist subinterval $[x_{i-1}, x_i]$ in $[a, b]$ such that $s in (x_{i-1}, x_i)$.



      Let $F(x_i) - F(x_{i-1}) lt frac{epsilon}{2mM}$. Let $M_i - m_i lt 2M$. Then $sum_{i=0}^{n}(M_i - m_i) lt 2mM$ because there are exactly $m$ terms in this sum, as there are exactly $m$ of the intervals in $[x_{i-1}, x_i]$.



      $U(P,f, F_s) - L(P,f, F_s) = (F(x_i)-F(x_{i-1}))sum_{i=0}^{n}(M_i - m_i) lt 2mM frac{epsilon}{2mM} = epsilon$.



      Therefore, $f in R(F_s)$.



      For the second part of the proof, we need to show: $int fdF_s = f(s)$.



      Am I doing this proof right so far? could someone help me with the rest?



      Thanks in advance.






      proof-verification metric-spaces riemann-integration







      share|cite|improve this question











      $endgroup$




      Let $a<s<b$ and let $f:[a, b] to mathbb{R}$ be a bounded function that is continous at the point s. Define



      $F_s(x) = begin{cases}
      0, & text{if $a le x lt s$} \
      1, & text{if $s le x le b$}
      end{cases}$



      Prove that $f in R(F_s)$ on the interval $[a, b]$ and that



      $int fdF_s = f(s)$.



      My attempt to the proof:



      Consider the partition,$P$ of the interval $[a, b]$ where $a=x_o le x_1 le ... le x_{i-1} le x_i le ... le x_n = b$.



      Let $M_i = supf(x) and m_i = inf f(x)$ for $x_{i-1} le x le x_i$.



      Given $epsilon gt 0$, and $P$. $fin R(F_s)$ if $U(P,f, F_s)-L(P,f,F_s) lt epsilon$.



      We know $F(x)$ is continuous at $s in [a,b]$. Let there exist subinterval $[x_{i-1}, x_i]$ in $[a, b]$ such that $s in (x_{i-1}, x_i)$.



      Let $F(x_i) - F(x_{i-1}) lt frac{epsilon}{2mM}$. Let $M_i - m_i lt 2M$. Then $sum_{i=0}^{n}(M_i - m_i) lt 2mM$ because there are exactly $m$ terms in this sum, as there are exactly $m$ of the intervals in $[x_{i-1}, x_i]$.



      $U(P,f, F_s) - L(P,f, F_s) = (F(x_i)-F(x_{i-1}))sum_{i=0}^{n}(M_i - m_i) lt 2mM frac{epsilon}{2mM} = epsilon$.



      Therefore, $f in R(F_s)$.



      For the second part of the proof, we need to show: $int fdF_s = f(s)$.



      Am I doing this proof right so far? could someone help me with the rest?



      Thanks in advance.






      proof-verification metric-spaces riemann-integration




      proof-verification metric-spaces riemann-integration






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 6 '18 at 22:34







      ISuckAtMathPleaseHELPME

















      asked Dec 6 '18 at 22:17









      ISuckAtMathPleaseHELPMEISuckAtMathPleaseHELPME

      1619




      1619






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          Everything looks good. You've proven that $f$ is Riemann-Lebesgue integrable with respect to $F_s$. Now pick any given sequence of partitions that allow the integral to converge $Pi_k$ and focus on the partition points just before and after $s$. Their contribution is $f(x_r)(F(x_r)-F(x_{r-1}))=f(x_r)cdot 1$. Now just use continuity of $f$ at $s$ and the definition of a sequence of partitions to conclude the integral converges to $f(s)$.






          share|cite|improve this answer









          $endgroup$













            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3029128%2fprove-that-f-in-rf-s-on-the-interval-a-b-and-that-int-f-df-s-fs%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            Everything looks good. You've proven that $f$ is Riemann-Lebesgue integrable with respect to $F_s$. Now pick any given sequence of partitions that allow the integral to converge $Pi_k$ and focus on the partition points just before and after $s$. Their contribution is $f(x_r)(F(x_r)-F(x_{r-1}))=f(x_r)cdot 1$. Now just use continuity of $f$ at $s$ and the definition of a sequence of partitions to conclude the integral converges to $f(s)$.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Everything looks good. You've proven that $f$ is Riemann-Lebesgue integrable with respect to $F_s$. Now pick any given sequence of partitions that allow the integral to converge $Pi_k$ and focus on the partition points just before and after $s$. Their contribution is $f(x_r)(F(x_r)-F(x_{r-1}))=f(x_r)cdot 1$. Now just use continuity of $f$ at $s$ and the definition of a sequence of partitions to conclude the integral converges to $f(s)$.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Everything looks good. You've proven that $f$ is Riemann-Lebesgue integrable with respect to $F_s$. Now pick any given sequence of partitions that allow the integral to converge $Pi_k$ and focus on the partition points just before and after $s$. Their contribution is $f(x_r)(F(x_r)-F(x_{r-1}))=f(x_r)cdot 1$. Now just use continuity of $f$ at $s$ and the definition of a sequence of partitions to conclude the integral converges to $f(s)$.






                share|cite|improve this answer









                $endgroup$



                Everything looks good. You've proven that $f$ is Riemann-Lebesgue integrable with respect to $F_s$. Now pick any given sequence of partitions that allow the integral to converge $Pi_k$ and focus on the partition points just before and after $s$. Their contribution is $f(x_r)(F(x_r)-F(x_{r-1}))=f(x_r)cdot 1$. Now just use continuity of $f$ at $s$ and the definition of a sequence of partitions to conclude the integral converges to $f(s)$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 6 '18 at 23:00









                Alex R.Alex R.

                25k12452




                25k12452






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3029128%2fprove-that-f-in-rf-s-on-the-interval-a-b-and-that-int-f-df-s-fs%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    -

                    Popular posts from this blog

                    How to change which sound is reproduced for terminal bell?

                    Image
                    3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
                    Read more

                    Can I use Tabulator js library in my java Spring + Thymeleaf project?

                    Image
                    0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
                    Read more

                    Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

                    Image
                    up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
                    Read more