munkres analysis integration question

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Let $[0,1]^2 = [0,1] times [0,1]$. Let $f: [0,1]^2 to mathbbR$ be defined by setting $f(x,y)=0$ if $y neq x$, and $f(x,y) = 1$ if $y=x$. Show that $f$ is integrable over $[0,1]^2$.






integration analysis measure-theory







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  • what kind of integral are you using? With the Lebesgue integral, the problem is trivial.
    – Stefan Smith
    Jan 11 '14 at 15:22














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Let $[0,1]^2 = [0,1] times [0,1]$. Let $f: [0,1]^2 to mathbbR$ be defined by setting $f(x,y)=0$ if $y neq x$, and $f(x,y) = 1$ if $y=x$. Show that $f$ is integrable over $[0,1]^2$.






integration analysis measure-theory







share|cite|improve this question























  • what kind of integral are you using? With the Lebesgue integral, the problem is trivial.
    – Stefan Smith
    Jan 11 '14 at 15:22












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Let $[0,1]^2 = [0,1] times [0,1]$. Let $f: [0,1]^2 to mathbbR$ be defined by setting $f(x,y)=0$ if $y neq x$, and $f(x,y) = 1$ if $y=x$. Show that $f$ is integrable over $[0,1]^2$.






integration analysis measure-theory







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Let $[0,1]^2 = [0,1] times [0,1]$. Let $f: [0,1]^2 to mathbbR$ be defined by setting $f(x,y)=0$ if $y neq x$, and $f(x,y) = 1$ if $y=x$. Show that $f$ is integrable over $[0,1]^2$.






integration analysis measure-theory




integration analysis measure-theory






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edited Jan 14 '14 at 18:21

























asked Jan 11 '14 at 13:38









user115722

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  • what kind of integral are you using? With the Lebesgue integral, the problem is trivial.
    – Stefan Smith
    Jan 11 '14 at 15:22
















  • what kind of integral are you using? With the Lebesgue integral, the problem is trivial.
    – Stefan Smith
    Jan 11 '14 at 15:22















what kind of integral are you using? With the Lebesgue integral, the problem is trivial.
– Stefan Smith
Jan 11 '14 at 15:22




what kind of integral are you using? With the Lebesgue integral, the problem is trivial.
– Stefan Smith
Jan 11 '14 at 15:22










2 Answers
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Given $ varepsilon> 0 $ is $ k in N $ such that $varepsilon >1/k$ consider a 'homogeneous' partition of $ [0,1] times [0,1] $, i.e., all sub- rectangles of partition has area $1/k^2$, we know that $ m_B = 0$ and $ m_B = 1$ if $B$ intection the line $ y = x $ is not empty set, so
$ S (f, P)-s (f, P) = sum_ B in P (m_B-m_B) vol (B) = sum_ B mbox such that intection the line $ y = x $ is not empty set = 1 / k <varepsilon $ hence f is integrable .






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    enter image description here



    Consider the partition $P$ like the above picture.



    $U(f, P) - L(f, P) = frac3 n - 2n^2$.



    $frac3 n - 2n^2 to 0 (n to infty)$.



    So, for any $epsilon > 0$, there exists a partition $P$ such that $U(f, P) - L(f, P) < epsilon$.



    By Theorem 10.3 (p.86), $f$ is integrable over $[0, 1] times [0, 1]$.






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      2 Answers
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      2 Answers
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      Given $ varepsilon> 0 $ is $ k in N $ such that $varepsilon >1/k$ consider a 'homogeneous' partition of $ [0,1] times [0,1] $, i.e., all sub- rectangles of partition has area $1/k^2$, we know that $ m_B = 0$ and $ m_B = 1$ if $B$ intection the line $ y = x $ is not empty set, so
      $ S (f, P)-s (f, P) = sum_ B in P (m_B-m_B) vol (B) = sum_ B mbox such that intection the line $ y = x $ is not empty set = 1 / k <varepsilon $ hence f is integrable .






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        Given $ varepsilon> 0 $ is $ k in N $ such that $varepsilon >1/k$ consider a 'homogeneous' partition of $ [0,1] times [0,1] $, i.e., all sub- rectangles of partition has area $1/k^2$, we know that $ m_B = 0$ and $ m_B = 1$ if $B$ intection the line $ y = x $ is not empty set, so
        $ S (f, P)-s (f, P) = sum_ B in P (m_B-m_B) vol (B) = sum_ B mbox such that intection the line $ y = x $ is not empty set = 1 / k <varepsilon $ hence f is integrable .






        share|cite|improve this answer
























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          Given $ varepsilon> 0 $ is $ k in N $ such that $varepsilon >1/k$ consider a 'homogeneous' partition of $ [0,1] times [0,1] $, i.e., all sub- rectangles of partition has area $1/k^2$, we know that $ m_B = 0$ and $ m_B = 1$ if $B$ intection the line $ y = x $ is not empty set, so
          $ S (f, P)-s (f, P) = sum_ B in P (m_B-m_B) vol (B) = sum_ B mbox such that intection the line $ y = x $ is not empty set = 1 / k <varepsilon $ hence f is integrable .






          share|cite|improve this answer














          Given $ varepsilon> 0 $ is $ k in N $ such that $varepsilon >1/k$ consider a 'homogeneous' partition of $ [0,1] times [0,1] $, i.e., all sub- rectangles of partition has area $1/k^2$, we know that $ m_B = 0$ and $ m_B = 1$ if $B$ intection the line $ y = x $ is not empty set, so
          $ S (f, P)-s (f, P) = sum_ B in P (m_B-m_B) vol (B) = sum_ B mbox such that intection the line $ y = x $ is not empty set = 1 / k <varepsilon $ hence f is integrable .







          share|cite|improve this answer














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          edited Apr 19 '14 at 4:30

























          answered Apr 19 '14 at 4:07









          Alcides de carvalho jr

          11311




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              enter image description here



              Consider the partition $P$ like the above picture.



              $U(f, P) - L(f, P) = frac3 n - 2n^2$.



              $frac3 n - 2n^2 to 0 (n to infty)$.



              So, for any $epsilon > 0$, there exists a partition $P$ such that $U(f, P) - L(f, P) < epsilon$.



              By Theorem 10.3 (p.86), $f$ is integrable over $[0, 1] times [0, 1]$.






              share|cite|improve this answer
























                up vote
                0
                down vote













                enter image description here



                Consider the partition $P$ like the above picture.



                $U(f, P) - L(f, P) = frac3 n - 2n^2$.



                $frac3 n - 2n^2 to 0 (n to infty)$.



                So, for any $epsilon > 0$, there exists a partition $P$ such that $U(f, P) - L(f, P) < epsilon$.



                By Theorem 10.3 (p.86), $f$ is integrable over $[0, 1] times [0, 1]$.






                share|cite|improve this answer






















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                  0
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                  up vote
                  0
                  down vote









                  enter image description here



                  Consider the partition $P$ like the above picture.



                  $U(f, P) - L(f, P) = frac3 n - 2n^2$.



                  $frac3 n - 2n^2 to 0 (n to infty)$.



                  So, for any $epsilon > 0$, there exists a partition $P$ such that $U(f, P) - L(f, P) < epsilon$.



                  By Theorem 10.3 (p.86), $f$ is integrable over $[0, 1] times [0, 1]$.






                  share|cite|improve this answer












                  enter image description here



                  Consider the partition $P$ like the above picture.



                  $U(f, P) - L(f, P) = frac3 n - 2n^2$.



                  $frac3 n - 2n^2 to 0 (n to infty)$.



                  So, for any $epsilon > 0$, there exists a partition $P$ such that $U(f, P) - L(f, P) < epsilon$.



                  By Theorem 10.3 (p.86), $f$ is integrable over $[0, 1] times [0, 1]$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Sep 3 at 9:25









                  tchappy ha

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