Prove that $f in R(F_s)$ on the interval $[a, b]$ and that $int f dF_s = f(s)$
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
asked Dec 6 at 22:17
ISuckAtMathPleaseHELPME
1549
add a comment |
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
asked Dec 6 at 22:17
ISuckAtMathPleaseHELPME
1549
add a comment |
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
asked Dec 6 at 22:17
ISuckAtMathPleaseHELPME
1549
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
asked Dec 6 at 22:17
ISuckAtMathPleaseHELPME
1549
asked Dec 6 at 22:17
ISuckAtMathPleaseHELPME
1549
edited Dec 6 at 22:34
asked Dec 6 at 22:17
ISuckAtMathPleaseHELPME
1549
asked Dec 6 at 22:17
ISuckAtMathPleaseHELPME
1549
asked Dec 6 at 22:17
ISuckAtMathPleaseHELPME
1549
1549
add a comment |
add a comment |
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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 $f(s)$.
answered Dec 6 at 23:00
Alex R.
24.7k12352
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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 $f(s)$.
answered Dec 6 at 23:00
Alex R.
24.7k12352
add a comment |
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 $f(s)$.
answered Dec 6 at 23:00
Alex R.
24.7k12352
add a comment |
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 $f(s)$.
answered Dec 6 at 23:00
Alex R.
24.7k12352
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 $f(s)$.
answered Dec 6 at 23:00
Alex R.
24.7k12352
answered Dec 6 at 23:00
Alex R.
24.7k12352
answered Dec 6 at 23:00
Alex R.
24.7k12352
answered Dec 6 at 23:00
Alex R.
24.7k12352
24.7k12352
add a comment |
add a comment |
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