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Riemann integrable or not?
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Let $(a_n)_1^infty$ be an increasing sequence in $(0,1)$ with limit $1$ and define $f:[0,1]tomathbbR$ such that $f(x)=1$ if $x=a_n$ for some $ninmathbbN$ and $f(x)=0$ otherwise. Is $f$ Riemann integrable on $[0,1]$?
I think it is Riemann integrable since $f$ is bounded and has a countable number of discontinuities, namely the $a_n$'s. But I don't know how to rigorously show it.
real-analysis
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Habagat Maliksi
10211
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up vote
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Let $(a_n)_1^infty$ be an increasing sequence in $(0,1)$ with limit $1$ and define $f:[0,1]tomathbbR$ such that $f(x)=1$ if $x=a_n$ for some $ninmathbbN$ and $f(x)=0$ otherwise. Is $f$ Riemann integrable on $[0,1]$?
I think it is Riemann integrable since $f$ is bounded and has a countable number of discontinuities, namely the $a_n$'s. But I don't know how to rigorously show it.
real-analysis
asked yesterday
Habagat Maliksi
10211
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $(a_n)_1^infty$ be an increasing sequence in $(0,1)$ with limit $1$ and define $f:[0,1]tomathbbR$ such that $f(x)=1$ if $x=a_n$ for some $ninmathbbN$ and $f(x)=0$ otherwise. Is $f$ Riemann integrable on $[0,1]$?
I think it is Riemann integrable since $f$ is bounded and has a countable number of discontinuities, namely the $a_n$'s. But I don't know how to rigorously show it.
real-analysis
asked yesterday
Habagat Maliksi
10211
Let $(a_n)_1^infty$ be an increasing sequence in $(0,1)$ with limit $1$ and define $f:[0,1]tomathbbR$ such that $f(x)=1$ if $x=a_n$ for some $ninmathbbN$ and $f(x)=0$ otherwise. Is $f$ Riemann integrable on $[0,1]$?
I think it is Riemann integrable since $f$ is bounded and has a countable number of discontinuities, namely the $a_n$'s. But I don't know how to rigorously show it.
real-analysis
asked yesterday
Habagat Maliksi
10211
asked yesterday
Habagat Maliksi
10211
asked yesterday
Habagat Maliksi
10211
asked yesterday
Habagat Maliksi
10211
10211
add a comment |Â
add a comment |Â
3 Answers
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Yes, it is integrable and its integral on $[0,1]$is $0$
There are only a countable set of discontinuity at your function which makes it integrable.
answered yesterday
Mohammad Riazi-Kermani
26.9k41849
so am i right sir?
– Habagat Maliksi
yesterday
Yes, of course. The function is integrable as you have indicated.
– Mohammad Riazi-Kermani
yesterday
add a comment |Â
up vote
0
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You are right about the Lebesgue criterion. Here is a proof of discontinuity.
Rigorous proof.
$blacktriangleleft$ Suppose $a_n$'s are mutually distinct [otherwise we could discard those terms with repeated values]. In other words $(a_n)$ is increasing strictly. Take $varepsilon_0 = 1/2$, then for each $a_n in [0,1]$, for all $delta colon 0< delta <mina_n+1-a_n, a_n - a_n-1$, there is $x = a_n+delta/2 in (a_n -delta, a_n+ delta)$ s.t. $$|f(x)-f(a_n)| =|0-1|=1 >1/2 =varepsilon _0.$$Hence $f$ is not continuous at $a_n blacktriangleright$.
answered yesterday
xbh
8906
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up vote
0
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Since the set
$$a_n $$
is countable, it is of measure zero, and it is clear that this set is also the set of discontinuities of $f$ (see @xbh's answer), hence by Lebesgue criterion, $f$ is Riemann integrable on [0,1].
answered yesterday
onurcanbektas
2,8991730
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Yes, it is integrable and its integral on $[0,1]$is $0$
There are only a countable set of discontinuity at your function which makes it integrable.
answered yesterday
Mohammad Riazi-Kermani
26.9k41849
so am i right sir?
– Habagat Maliksi
yesterday
Yes, of course. The function is integrable as you have indicated.
– Mohammad Riazi-Kermani
yesterday
add a comment |Â
up vote
0
down vote
Yes, it is integrable and its integral on $[0,1]$is $0$
There are only a countable set of discontinuity at your function which makes it integrable.
answered yesterday
Mohammad Riazi-Kermani
26.9k41849
so am i right sir?
– Habagat Maliksi
yesterday
Yes, of course. The function is integrable as you have indicated.
– Mohammad Riazi-Kermani
yesterday
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Yes, it is integrable and its integral on $[0,1]$is $0$
There are only a countable set of discontinuity at your function which makes it integrable.
answered yesterday
Mohammad Riazi-Kermani
26.9k41849
Yes, it is integrable and its integral on $[0,1]$is $0$
There are only a countable set of discontinuity at your function which makes it integrable.
answered yesterday
Mohammad Riazi-Kermani
26.9k41849
answered yesterday
Mohammad Riazi-Kermani
26.9k41849
answered yesterday
Mohammad Riazi-Kermani
26.9k41849
answered yesterday
Mohammad Riazi-Kermani
26.9k41849
26.9k41849
so am i right sir?
– Habagat Maliksi
yesterday
Yes, of course. The function is integrable as you have indicated.
– Mohammad Riazi-Kermani
yesterday
add a comment |Â
so am i right sir?
– Habagat Maliksi
yesterday
Yes, of course. The function is integrable as you have indicated.
– Mohammad Riazi-Kermani
yesterday
so am i right sir?
– Habagat Maliksi
yesterday
so am i right sir?
– Habagat Maliksi
yesterday
Yes, of course. The function is integrable as you have indicated.
– Mohammad Riazi-Kermani
yesterday
Yes, of course. The function is integrable as you have indicated.
– Mohammad Riazi-Kermani
yesterday
add a comment |Â
up vote
0
down vote
You are right about the Lebesgue criterion. Here is a proof of discontinuity.
Rigorous proof.
$blacktriangleleft$ Suppose $a_n$'s are mutually distinct [otherwise we could discard those terms with repeated values]. In other words $(a_n)$ is increasing strictly. Take $varepsilon_0 = 1/2$, then for each $a_n in [0,1]$, for all $delta colon 0< delta <mina_n+1-a_n, a_n - a_n-1$, there is $x = a_n+delta/2 in (a_n -delta, a_n+ delta)$ s.t. $$|f(x)-f(a_n)| =|0-1|=1 >1/2 =varepsilon _0.$$Hence $f$ is not continuous at $a_n blacktriangleright$.
answered yesterday
xbh
8906
add a comment |Â
up vote
0
down vote
You are right about the Lebesgue criterion. Here is a proof of discontinuity.
Rigorous proof.
$blacktriangleleft$ Suppose $a_n$'s are mutually distinct [otherwise we could discard those terms with repeated values]. In other words $(a_n)$ is increasing strictly. Take $varepsilon_0 = 1/2$, then for each $a_n in [0,1]$, for all $delta colon 0< delta <mina_n+1-a_n, a_n - a_n-1$, there is $x = a_n+delta/2 in (a_n -delta, a_n+ delta)$ s.t. $$|f(x)-f(a_n)| =|0-1|=1 >1/2 =varepsilon _0.$$Hence $f$ is not continuous at $a_n blacktriangleright$.
answered yesterday
xbh
8906
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You are right about the Lebesgue criterion. Here is a proof of discontinuity.
Rigorous proof.
$blacktriangleleft$ Suppose $a_n$'s are mutually distinct [otherwise we could discard those terms with repeated values]. In other words $(a_n)$ is increasing strictly. Take $varepsilon_0 = 1/2$, then for each $a_n in [0,1]$, for all $delta colon 0< delta <mina_n+1-a_n, a_n - a_n-1$, there is $x = a_n+delta/2 in (a_n -delta, a_n+ delta)$ s.t. $$|f(x)-f(a_n)| =|0-1|=1 >1/2 =varepsilon _0.$$Hence $f$ is not continuous at $a_n blacktriangleright$.
answered yesterday
xbh
8906
You are right about the Lebesgue criterion. Here is a proof of discontinuity.
Rigorous proof.
$blacktriangleleft$ Suppose $a_n$'s are mutually distinct [otherwise we could discard those terms with repeated values]. In other words $(a_n)$ is increasing strictly. Take $varepsilon_0 = 1/2$, then for each $a_n in [0,1]$, for all $delta colon 0< delta <mina_n+1-a_n, a_n - a_n-1$, there is $x = a_n+delta/2 in (a_n -delta, a_n+ delta)$ s.t. $$|f(x)-f(a_n)| =|0-1|=1 >1/2 =varepsilon _0.$$Hence $f$ is not continuous at $a_n blacktriangleright$.
answered yesterday
xbh
8906
answered yesterday
xbh
8906
answered yesterday
xbh
8906
answered yesterday
xbh
8906
8906
add a comment |Â
add a comment |Â
up vote
0
down vote
Since the set
$$a_n $$
is countable, it is of measure zero, and it is clear that this set is also the set of discontinuities of $f$ (see @xbh's answer), hence by Lebesgue criterion, $f$ is Riemann integrable on [0,1].
answered yesterday
onurcanbektas
2,8991730
add a comment |Â
up vote
0
down vote
Since the set
$$a_n $$
is countable, it is of measure zero, and it is clear that this set is also the set of discontinuities of $f$ (see @xbh's answer), hence by Lebesgue criterion, $f$ is Riemann integrable on [0,1].
answered yesterday
onurcanbektas
2,8991730
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Since the set
$$a_n $$
is countable, it is of measure zero, and it is clear that this set is also the set of discontinuities of $f$ (see @xbh's answer), hence by Lebesgue criterion, $f$ is Riemann integrable on [0,1].
answered yesterday
onurcanbektas
2,8991730
Since the set
$$a_n $$
is countable, it is of measure zero, and it is clear that this set is also the set of discontinuities of $f$ (see @xbh's answer), hence by Lebesgue criterion, $f$ is Riemann integrable on [0,1].
answered yesterday
onurcanbektas
2,8991730
answered yesterday
onurcanbektas
2,8991730
answered yesterday
onurcanbektas
2,8991730
answered yesterday
onurcanbektas
2,8991730
2,8991730
add a comment |Â
add a comment |Â
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