In a measurable partition of an interval, the sum of the measures of the subsets in the partition equals the...











up vote
1
down vote

favorite












I am working through A User-Friendly Introduction to Lebesgue Measure and Integration, by Gail S. Nelson. On page 67 Nelson defines a measurable partition of the interval $[a,b]$ to be finite collection $ {E_j }_{j=1}^{n}$ of subsets of $[a,b]$ such that: (1) $E_j$ is measurable for each $j$, (2) $bigcup_{j=1}^{n}E_j=[a,b]$, and (3) $m(E_i cap E_j)=0$ whenever $inot=j$. On page 68 she then remarks that given any measurable partition of $[a,b]$, we can convert it into a pairwise disjoint measurable partition like so: set $F_1=E_1$, and $F_j=E_j setminus(bigcup_{i=1}^{j-1}E_i)$ for $j=2,3,...n$. I can see why these sets are pairwise disjoint, and I can see how each one is measurable, but I'm unable to prove that (1) their union is $[a,b]$ and (2) that $m(E_j)=m(F_j)$ for each $j=2,3,...n$. I suspect I'm overlooking something simple. Any help is greatly appreciated!










share|cite|improve this question


























    up vote
    1
    down vote

    favorite












    I am working through A User-Friendly Introduction to Lebesgue Measure and Integration, by Gail S. Nelson. On page 67 Nelson defines a measurable partition of the interval $[a,b]$ to be finite collection $ {E_j }_{j=1}^{n}$ of subsets of $[a,b]$ such that: (1) $E_j$ is measurable for each $j$, (2) $bigcup_{j=1}^{n}E_j=[a,b]$, and (3) $m(E_i cap E_j)=0$ whenever $inot=j$. On page 68 she then remarks that given any measurable partition of $[a,b]$, we can convert it into a pairwise disjoint measurable partition like so: set $F_1=E_1$, and $F_j=E_j setminus(bigcup_{i=1}^{j-1}E_i)$ for $j=2,3,...n$. I can see why these sets are pairwise disjoint, and I can see how each one is measurable, but I'm unable to prove that (1) their union is $[a,b]$ and (2) that $m(E_j)=m(F_j)$ for each $j=2,3,...n$. I suspect I'm overlooking something simple. Any help is greatly appreciated!










    share|cite|improve this question
























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I am working through A User-Friendly Introduction to Lebesgue Measure and Integration, by Gail S. Nelson. On page 67 Nelson defines a measurable partition of the interval $[a,b]$ to be finite collection $ {E_j }_{j=1}^{n}$ of subsets of $[a,b]$ such that: (1) $E_j$ is measurable for each $j$, (2) $bigcup_{j=1}^{n}E_j=[a,b]$, and (3) $m(E_i cap E_j)=0$ whenever $inot=j$. On page 68 she then remarks that given any measurable partition of $[a,b]$, we can convert it into a pairwise disjoint measurable partition like so: set $F_1=E_1$, and $F_j=E_j setminus(bigcup_{i=1}^{j-1}E_i)$ for $j=2,3,...n$. I can see why these sets are pairwise disjoint, and I can see how each one is measurable, but I'm unable to prove that (1) their union is $[a,b]$ and (2) that $m(E_j)=m(F_j)$ for each $j=2,3,...n$. I suspect I'm overlooking something simple. Any help is greatly appreciated!










      share|cite|improve this question













      I am working through A User-Friendly Introduction to Lebesgue Measure and Integration, by Gail S. Nelson. On page 67 Nelson defines a measurable partition of the interval $[a,b]$ to be finite collection $ {E_j }_{j=1}^{n}$ of subsets of $[a,b]$ such that: (1) $E_j$ is measurable for each $j$, (2) $bigcup_{j=1}^{n}E_j=[a,b]$, and (3) $m(E_i cap E_j)=0$ whenever $inot=j$. On page 68 she then remarks that given any measurable partition of $[a,b]$, we can convert it into a pairwise disjoint measurable partition like so: set $F_1=E_1$, and $F_j=E_j setminus(bigcup_{i=1}^{j-1}E_i)$ for $j=2,3,...n$. I can see why these sets are pairwise disjoint, and I can see how each one is measurable, but I'm unable to prove that (1) their union is $[a,b]$ and (2) that $m(E_j)=m(F_j)$ for each $j=2,3,...n$. I suspect I'm overlooking something simple. Any help is greatly appreciated!







      real-analysis lebesgue-integral lebesgue-measure






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 15 at 22:50









      Alex Sanger

      624




      624






















          1 Answer
          1






          active

          oldest

          votes

















          up vote
          1
          down vote



          accepted










          Let's start with $(1)$. Since $F_j subset E_j subset [a,b]$, we only have to see that $[a,b] subset bigcup_j F_j$. Take $x in [a,b]$ and let $s = min{n : x in E_j}$. This is well defined because there exists some $j$ for which $x in E_j$ since the $E_j$s partition $[a,b]$. By construction, $x in E_s$ but necessarily, $x not in E_t$ when $t < s$. Hence,



          $$
          x in E_s setminus bigcup_{t < s}E_t = E_s setminus bigcup_{t=1}^{s-1}E_t = F_s.
          $$


          This proves that, in effect, $[a,b]$ is the disjoint union of $F_1, dots, F_n$.



          Finally, note that for each $j in [n]$,



          $$
          F_j = E_j setminus bigcup_{i=1}^{j-1}E_i = E_j setminus bigcup_{i=1}^{j-1}E_j cap E_i
          $$



          and so since everything here is measurable and of finite measure,



          $$
          |F_j| = |E_j| - left|bigcup_{i < j}E_j cap E_iright|. tag{$star$}
          $$



          However, by hypothesis we have that



          $$
          left|bigcup_{i < j}E_j cap E_iright| leq sum_{i < j}|E_j cap E_i| = 0,
          $$



          and so plugging that in $(star)$ proves that, in effect,



          $$
          |F_j| = |E_j|.
          $$



          The intuition here is that $F_i$ is $E_i$ except removing some of the intersections with other $E_j$, but since all of these are of zero measure by hypothesis, $F_i$ and $E_i$ have the same measure: the sets we have 'cropped' are (measure wise) negligible.






          share|cite|improve this answer





















            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',
            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%2f3000444%2fin-a-measurable-partition-of-an-interval-the-sum-of-the-measures-of-the-subsets%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








            up vote
            1
            down vote



            accepted










            Let's start with $(1)$. Since $F_j subset E_j subset [a,b]$, we only have to see that $[a,b] subset bigcup_j F_j$. Take $x in [a,b]$ and let $s = min{n : x in E_j}$. This is well defined because there exists some $j$ for which $x in E_j$ since the $E_j$s partition $[a,b]$. By construction, $x in E_s$ but necessarily, $x not in E_t$ when $t < s$. Hence,



            $$
            x in E_s setminus bigcup_{t < s}E_t = E_s setminus bigcup_{t=1}^{s-1}E_t = F_s.
            $$


            This proves that, in effect, $[a,b]$ is the disjoint union of $F_1, dots, F_n$.



            Finally, note that for each $j in [n]$,



            $$
            F_j = E_j setminus bigcup_{i=1}^{j-1}E_i = E_j setminus bigcup_{i=1}^{j-1}E_j cap E_i
            $$



            and so since everything here is measurable and of finite measure,



            $$
            |F_j| = |E_j| - left|bigcup_{i < j}E_j cap E_iright|. tag{$star$}
            $$



            However, by hypothesis we have that



            $$
            left|bigcup_{i < j}E_j cap E_iright| leq sum_{i < j}|E_j cap E_i| = 0,
            $$



            and so plugging that in $(star)$ proves that, in effect,



            $$
            |F_j| = |E_j|.
            $$



            The intuition here is that $F_i$ is $E_i$ except removing some of the intersections with other $E_j$, but since all of these are of zero measure by hypothesis, $F_i$ and $E_i$ have the same measure: the sets we have 'cropped' are (measure wise) negligible.






            share|cite|improve this answer

























              up vote
              1
              down vote



              accepted










              Let's start with $(1)$. Since $F_j subset E_j subset [a,b]$, we only have to see that $[a,b] subset bigcup_j F_j$. Take $x in [a,b]$ and let $s = min{n : x in E_j}$. This is well defined because there exists some $j$ for which $x in E_j$ since the $E_j$s partition $[a,b]$. By construction, $x in E_s$ but necessarily, $x not in E_t$ when $t < s$. Hence,



              $$
              x in E_s setminus bigcup_{t < s}E_t = E_s setminus bigcup_{t=1}^{s-1}E_t = F_s.
              $$


              This proves that, in effect, $[a,b]$ is the disjoint union of $F_1, dots, F_n$.



              Finally, note that for each $j in [n]$,



              $$
              F_j = E_j setminus bigcup_{i=1}^{j-1}E_i = E_j setminus bigcup_{i=1}^{j-1}E_j cap E_i
              $$



              and so since everything here is measurable and of finite measure,



              $$
              |F_j| = |E_j| - left|bigcup_{i < j}E_j cap E_iright|. tag{$star$}
              $$



              However, by hypothesis we have that



              $$
              left|bigcup_{i < j}E_j cap E_iright| leq sum_{i < j}|E_j cap E_i| = 0,
              $$



              and so plugging that in $(star)$ proves that, in effect,



              $$
              |F_j| = |E_j|.
              $$



              The intuition here is that $F_i$ is $E_i$ except removing some of the intersections with other $E_j$, but since all of these are of zero measure by hypothesis, $F_i$ and $E_i$ have the same measure: the sets we have 'cropped' are (measure wise) negligible.






              share|cite|improve this answer























                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                Let's start with $(1)$. Since $F_j subset E_j subset [a,b]$, we only have to see that $[a,b] subset bigcup_j F_j$. Take $x in [a,b]$ and let $s = min{n : x in E_j}$. This is well defined because there exists some $j$ for which $x in E_j$ since the $E_j$s partition $[a,b]$. By construction, $x in E_s$ but necessarily, $x not in E_t$ when $t < s$. Hence,



                $$
                x in E_s setminus bigcup_{t < s}E_t = E_s setminus bigcup_{t=1}^{s-1}E_t = F_s.
                $$


                This proves that, in effect, $[a,b]$ is the disjoint union of $F_1, dots, F_n$.



                Finally, note that for each $j in [n]$,



                $$
                F_j = E_j setminus bigcup_{i=1}^{j-1}E_i = E_j setminus bigcup_{i=1}^{j-1}E_j cap E_i
                $$



                and so since everything here is measurable and of finite measure,



                $$
                |F_j| = |E_j| - left|bigcup_{i < j}E_j cap E_iright|. tag{$star$}
                $$



                However, by hypothesis we have that



                $$
                left|bigcup_{i < j}E_j cap E_iright| leq sum_{i < j}|E_j cap E_i| = 0,
                $$



                and so plugging that in $(star)$ proves that, in effect,



                $$
                |F_j| = |E_j|.
                $$



                The intuition here is that $F_i$ is $E_i$ except removing some of the intersections with other $E_j$, but since all of these are of zero measure by hypothesis, $F_i$ and $E_i$ have the same measure: the sets we have 'cropped' are (measure wise) negligible.






                share|cite|improve this answer












                Let's start with $(1)$. Since $F_j subset E_j subset [a,b]$, we only have to see that $[a,b] subset bigcup_j F_j$. Take $x in [a,b]$ and let $s = min{n : x in E_j}$. This is well defined because there exists some $j$ for which $x in E_j$ since the $E_j$s partition $[a,b]$. By construction, $x in E_s$ but necessarily, $x not in E_t$ when $t < s$. Hence,



                $$
                x in E_s setminus bigcup_{t < s}E_t = E_s setminus bigcup_{t=1}^{s-1}E_t = F_s.
                $$


                This proves that, in effect, $[a,b]$ is the disjoint union of $F_1, dots, F_n$.



                Finally, note that for each $j in [n]$,



                $$
                F_j = E_j setminus bigcup_{i=1}^{j-1}E_i = E_j setminus bigcup_{i=1}^{j-1}E_j cap E_i
                $$



                and so since everything here is measurable and of finite measure,



                $$
                |F_j| = |E_j| - left|bigcup_{i < j}E_j cap E_iright|. tag{$star$}
                $$



                However, by hypothesis we have that



                $$
                left|bigcup_{i < j}E_j cap E_iright| leq sum_{i < j}|E_j cap E_i| = 0,
                $$



                and so plugging that in $(star)$ proves that, in effect,



                $$
                |F_j| = |E_j|.
                $$



                The intuition here is that $F_i$ is $E_i$ except removing some of the intersections with other $E_j$, but since all of these are of zero measure by hypothesis, $F_i$ and $E_i$ have the same measure: the sets we have 'cropped' are (measure wise) negligible.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 15 at 23:10









                Guido A.

                6,6321730




                6,6321730






























                     

                    draft saved


                    draft discarded



















































                     


                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3000444%2fin-a-measurable-partition-of-an-interval-the-sum-of-the-measures-of-the-subsets%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

                    AnyDesk - Fatal Program Failure

                    How to calibrate 16:9 built-in touch-screen to a 4:3 resolution?

                    QoS: MAC-Priority for clients behind a repeater