Tricky analysis proof involving the sum of integrals
$begingroup$
The question is as follows:
Let $P:a=t_{0}<t_{1}<...<t_{m}=b $ be a partition of $[a,b]$, $f_{j}$ be a Riemann integrable function on $[t_{j-1},t_{j}] forall j=1,...,m$, and let $f:[a,b]rightarrow mathbb{R}$ be a function with the property
$$frestriction_{(t_{j-1},t_{j})}=f_jrestriction_{(t_{j-1},t_{j})} forall j=1,...m.$$
Prove that
$$int_{a}^{b}f(x) dx=sum_{j=1}^{m}int_{t_{j-1}}^{t_{j}}f_{j}(x)dx.$$
I already know that $f$ is Riemann integrable iff $frestriction_{(t_{j-1},t_{j})}$ is Riemann integrable for each $j=1,...m.$ I have also proved this, so I believe the proof just needs to be shown for the case $m=1.$
Does anyone know how to tackle this?
Thanks in advance!
real-analysis integration riemann-integration
asked Dec 8 '18 at 19:59
MathIsLife12MathIsLife12
613211
$endgroup$
add a comment |
$begingroup$
The question is as follows:
Let $P:a=t_{0}<t_{1}<...<t_{m}=b $ be a partition of $[a,b]$, $f_{j}$ be a Riemann integrable function on $[t_{j-1},t_{j}] forall j=1,...,m$, and let $f:[a,b]rightarrow mathbb{R}$ be a function with the property
$$frestriction_{(t_{j-1},t_{j})}=f_jrestriction_{(t_{j-1},t_{j})} forall j=1,...m.$$
Prove that
$$int_{a}^{b}f(x) dx=sum_{j=1}^{m}int_{t_{j-1}}^{t_{j}}f_{j}(x)dx.$$
I already know that $f$ is Riemann integrable iff $frestriction_{(t_{j-1},t_{j})}$ is Riemann integrable for each $j=1,...m.$ I have also proved this, so I believe the proof just needs to be shown for the case $m=1.$
Does anyone know how to tackle this?
Thanks in advance!
real-analysis integration riemann-integration
asked Dec 8 '18 at 19:59
MathIsLife12MathIsLife12
613211
$endgroup$
$begingroup$
There is absolutely nothing to prove when $m=1$ because $f_1=f$ in that case.
$endgroup$
– Kavi Rama Murthy
Dec 9 '18 at 0:07
$begingroup$
@KaviRamaMurthy What is the correct approach then?
$endgroup$
– MathIsLife12
Dec 9 '18 at 1:35
add a comment |
$begingroup$
The question is as follows:
Let $P:a=t_{0}<t_{1}<...<t_{m}=b $ be a partition of $[a,b]$, $f_{j}$ be a Riemann integrable function on $[t_{j-1},t_{j}] forall j=1,...,m$, and let $f:[a,b]rightarrow mathbb{R}$ be a function with the property
$$frestriction_{(t_{j-1},t_{j})}=f_jrestriction_{(t_{j-1},t_{j})} forall j=1,...m.$$
Prove that
$$int_{a}^{b}f(x) dx=sum_{j=1}^{m}int_{t_{j-1}}^{t_{j}}f_{j}(x)dx.$$
I already know that $f$ is Riemann integrable iff $frestriction_{(t_{j-1},t_{j})}$ is Riemann integrable for each $j=1,...m.$ I have also proved this, so I believe the proof just needs to be shown for the case $m=1.$
Does anyone know how to tackle this?
Thanks in advance!
real-analysis integration riemann-integration
asked Dec 8 '18 at 19:59
MathIsLife12MathIsLife12
613211
$endgroup$
The question is as follows:
Let $P:a=t_{0}<t_{1}<...<t_{m}=b $ be a partition of $[a,b]$, $f_{j}$ be a Riemann integrable function on $[t_{j-1},t_{j}] forall j=1,...,m$, and let $f:[a,b]rightarrow mathbb{R}$ be a function with the property
$$frestriction_{(t_{j-1},t_{j})}=f_jrestriction_{(t_{j-1},t_{j})} forall j=1,...m.$$
Prove that
$$int_{a}^{b}f(x) dx=sum_{j=1}^{m}int_{t_{j-1}}^{t_{j}}f_{j}(x)dx.$$
I already know that $f$ is Riemann integrable iff $frestriction_{(t_{j-1},t_{j})}$ is Riemann integrable for each $j=1,...m.$ I have also proved this, so I believe the proof just needs to be shown for the case $m=1.$
Does anyone know how to tackle this?
Thanks in advance!
real-analysis integration riemann-integration
real-analysis integration riemann-integration
asked Dec 8 '18 at 19:59
MathIsLife12MathIsLife12
613211
asked Dec 8 '18 at 19:59
MathIsLife12MathIsLife12
613211
asked Dec 8 '18 at 19:59
MathIsLife12MathIsLife12
613211
asked Dec 8 '18 at 19:59
MathIsLife12MathIsLife12
613211
asked Dec 8 '18 at 19:59
MathIsLife12MathIsLife12
613211
613211
$begingroup$
There is absolutely nothing to prove when $m=1$ because $f_1=f$ in that case.
$endgroup$
– Kavi Rama Murthy
Dec 9 '18 at 0:07
$begingroup$
@KaviRamaMurthy What is the correct approach then?
$endgroup$
– MathIsLife12
Dec 9 '18 at 1:35
add a comment |
$begingroup$
There is absolutely nothing to prove when $m=1$ because $f_1=f$ in that case.
$endgroup$
– Kavi Rama Murthy
Dec 9 '18 at 0:07
$begingroup$
@KaviRamaMurthy What is the correct approach then?
$endgroup$
– MathIsLife12
Dec 9 '18 at 1:35
$begingroup$
There is absolutely nothing to prove when $m=1$ because $f_1=f$ in that case.
$endgroup$
– Kavi Rama Murthy
Dec 9 '18 at 0:07
$begingroup$
There is absolutely nothing to prove when $m=1$ because $f_1=f$ in that case.
$endgroup$
– Kavi Rama Murthy
Dec 9 '18 at 0:07
$begingroup$
@KaviRamaMurthy What is the correct approach then?
$endgroup$
– MathIsLife12
Dec 9 '18 at 1:35
$begingroup$
@KaviRamaMurthy What is the correct approach then?
$endgroup$
– MathIsLife12
Dec 9 '18 at 1:35
add a comment |
1 Answer
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$begingroup$
This is just a generalization of integrability of $f$ on $[a,c]$ and $[c,b]$ implies integrability on $[a,b]$ and
$$int_a^b f(x) , dx = int_a^c f(x) , dx + int_c^b f(x) , dx $$
There exist partitions $P_j$ of $[t_{j-1},t_j]$ such that for $j=1,ldots,m$, upper and lower Darboux sums satisfy
$$U(P_j,f_j) - L(P_j,f_j) < frac{epsilon}{m}$$
With the partition $P = bigcup_{j=1}^n P_j$ we have
$$U(P,f) - L(P,f) = sum_{j=1}^n [ U(P_j,f_j) - L(P_j,f_j)] < epsilon$$
and $f$ is integrable by the Riemann criterion.
We must also have
$$L(P,f) leqslant int_a^b f(x) , dx leqslant U(P,f), \ L(P,f) = sum_{j=1}^n L(P_j,f_j) = sum_{j=1}^n int_{t_{j-1}}^{t_j} f_j(x) , dx leqslant sum_{j=1}^n U(P_j,f_j) = U(P,f)$$
Thus,
$$-epsilon < -[U(P,f) - L(P,f)] < int_a^b f(x) , dx - sum_{j=1}^n int_{t_{j-1}}^{t_j} f_j(x) , dx < U(P,f) - L(P,f) < epsilon,$$
which implies (since $epsilon$ can be chosen arbitrarily close to $0$)
$$int_a^b f(x) , dx = sum_{j=1}^n int_{t_{j-1}}^{t_j} f_j(x) , dx $$
answered Dec 9 '18 at 6:38
RRLRRL
52.7k42573
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add a comment |
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1 Answer
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$begingroup$
This is just a generalization of integrability of $f$ on $[a,c]$ and $[c,b]$ implies integrability on $[a,b]$ and
$$int_a^b f(x) , dx = int_a^c f(x) , dx + int_c^b f(x) , dx $$
There exist partitions $P_j$ of $[t_{j-1},t_j]$ such that for $j=1,ldots,m$, upper and lower Darboux sums satisfy
$$U(P_j,f_j) - L(P_j,f_j) < frac{epsilon}{m}$$
With the partition $P = bigcup_{j=1}^n P_j$ we have
$$U(P,f) - L(P,f) = sum_{j=1}^n [ U(P_j,f_j) - L(P_j,f_j)] < epsilon$$
and $f$ is integrable by the Riemann criterion.
We must also have
$$L(P,f) leqslant int_a^b f(x) , dx leqslant U(P,f), \ L(P,f) = sum_{j=1}^n L(P_j,f_j) = sum_{j=1}^n int_{t_{j-1}}^{t_j} f_j(x) , dx leqslant sum_{j=1}^n U(P_j,f_j) = U(P,f)$$
Thus,
$$-epsilon < -[U(P,f) - L(P,f)] < int_a^b f(x) , dx - sum_{j=1}^n int_{t_{j-1}}^{t_j} f_j(x) , dx < U(P,f) - L(P,f) < epsilon,$$
which implies (since $epsilon$ can be chosen arbitrarily close to $0$)
$$int_a^b f(x) , dx = sum_{j=1}^n int_{t_{j-1}}^{t_j} f_j(x) , dx $$
answered Dec 9 '18 at 6:38
RRLRRL
52.7k42573
$endgroup$
add a comment |
$begingroup$
This is just a generalization of integrability of $f$ on $[a,c]$ and $[c,b]$ implies integrability on $[a,b]$ and
$$int_a^b f(x) , dx = int_a^c f(x) , dx + int_c^b f(x) , dx $$
There exist partitions $P_j$ of $[t_{j-1},t_j]$ such that for $j=1,ldots,m$, upper and lower Darboux sums satisfy
$$U(P_j,f_j) - L(P_j,f_j) < frac{epsilon}{m}$$
With the partition $P = bigcup_{j=1}^n P_j$ we have
$$U(P,f) - L(P,f) = sum_{j=1}^n [ U(P_j,f_j) - L(P_j,f_j)] < epsilon$$
and $f$ is integrable by the Riemann criterion.
We must also have
$$L(P,f) leqslant int_a^b f(x) , dx leqslant U(P,f), \ L(P,f) = sum_{j=1}^n L(P_j,f_j) = sum_{j=1}^n int_{t_{j-1}}^{t_j} f_j(x) , dx leqslant sum_{j=1}^n U(P_j,f_j) = U(P,f)$$
Thus,
$$-epsilon < -[U(P,f) - L(P,f)] < int_a^b f(x) , dx - sum_{j=1}^n int_{t_{j-1}}^{t_j} f_j(x) , dx < U(P,f) - L(P,f) < epsilon,$$
which implies (since $epsilon$ can be chosen arbitrarily close to $0$)
$$int_a^b f(x) , dx = sum_{j=1}^n int_{t_{j-1}}^{t_j} f_j(x) , dx $$
answered Dec 9 '18 at 6:38
RRLRRL
52.7k42573
$endgroup$
add a comment |
$begingroup$
This is just a generalization of integrability of $f$ on $[a,c]$ and $[c,b]$ implies integrability on $[a,b]$ and
$$int_a^b f(x) , dx = int_a^c f(x) , dx + int_c^b f(x) , dx $$
There exist partitions $P_j$ of $[t_{j-1},t_j]$ such that for $j=1,ldots,m$, upper and lower Darboux sums satisfy
$$U(P_j,f_j) - L(P_j,f_j) < frac{epsilon}{m}$$
With the partition $P = bigcup_{j=1}^n P_j$ we have
$$U(P,f) - L(P,f) = sum_{j=1}^n [ U(P_j,f_j) - L(P_j,f_j)] < epsilon$$
and $f$ is integrable by the Riemann criterion.
We must also have
$$L(P,f) leqslant int_a^b f(x) , dx leqslant U(P,f), \ L(P,f) = sum_{j=1}^n L(P_j,f_j) = sum_{j=1}^n int_{t_{j-1}}^{t_j} f_j(x) , dx leqslant sum_{j=1}^n U(P_j,f_j) = U(P,f)$$
Thus,
$$-epsilon < -[U(P,f) - L(P,f)] < int_a^b f(x) , dx - sum_{j=1}^n int_{t_{j-1}}^{t_j} f_j(x) , dx < U(P,f) - L(P,f) < epsilon,$$
which implies (since $epsilon$ can be chosen arbitrarily close to $0$)
$$int_a^b f(x) , dx = sum_{j=1}^n int_{t_{j-1}}^{t_j} f_j(x) , dx $$
answered Dec 9 '18 at 6:38
RRLRRL
52.7k42573
$endgroup$
This is just a generalization of integrability of $f$ on $[a,c]$ and $[c,b]$ implies integrability on $[a,b]$ and
$$int_a^b f(x) , dx = int_a^c f(x) , dx + int_c^b f(x) , dx $$
There exist partitions $P_j$ of $[t_{j-1},t_j]$ such that for $j=1,ldots,m$, upper and lower Darboux sums satisfy
$$U(P_j,f_j) - L(P_j,f_j) < frac{epsilon}{m}$$
With the partition $P = bigcup_{j=1}^n P_j$ we have
$$U(P,f) - L(P,f) = sum_{j=1}^n [ U(P_j,f_j) - L(P_j,f_j)] < epsilon$$
and $f$ is integrable by the Riemann criterion.
We must also have
$$L(P,f) leqslant int_a^b f(x) , dx leqslant U(P,f), \ L(P,f) = sum_{j=1}^n L(P_j,f_j) = sum_{j=1}^n int_{t_{j-1}}^{t_j} f_j(x) , dx leqslant sum_{j=1}^n U(P_j,f_j) = U(P,f)$$
Thus,
$$-epsilon < -[U(P,f) - L(P,f)] < int_a^b f(x) , dx - sum_{j=1}^n int_{t_{j-1}}^{t_j} f_j(x) , dx < U(P,f) - L(P,f) < epsilon,$$
which implies (since $epsilon$ can be chosen arbitrarily close to $0$)
$$int_a^b f(x) , dx = sum_{j=1}^n int_{t_{j-1}}^{t_j} f_j(x) , dx $$
answered Dec 9 '18 at 6:38
RRLRRL
52.7k42573
answered Dec 9 '18 at 6:38
RRLRRL
52.7k42573
answered Dec 9 '18 at 6:38
RRLRRL
52.7k42573
answered Dec 9 '18 at 6:38
RRLRRL
52.7k42573
52.7k42573
add a comment |
add a comment |
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$begingroup$
There is absolutely nothing to prove when $m=1$ because $f_1=f$ in that case.
$endgroup$
– Kavi Rama Murthy
Dec 9 '18 at 0:07
$begingroup$
@KaviRamaMurthy What is the correct approach then?
$endgroup$
– MathIsLife12
Dec 9 '18 at 1:35