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












0














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







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    0














    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



























      0












      0








      0







      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















      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






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      edited Dec 6 at 22:34

























      asked Dec 6 at 22:17









      ISuckAtMathPleaseHELPME

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          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 $f(s)$.






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            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 $f(s)$.






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              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
























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                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












                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 $f(s)$.







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                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 6 at 23:00









                Alex R.

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