Mixing
Is $f$ and $g$ is Riemann-integrable?
let $f : [0,1] rightarrow mathbb{R}$ and $g:[0,1] rightarrow mathbb{R}$ be two function define by
$$f(x) =
begin{cases}
frac {1}{n},text{ if }x = frac{1}{n},n in mathbb{N}\
0, text{otherwise}end{cases}$$ and $$g(x) =
begin{cases}
n,text{ if } x = frac{1}{n}, n in mathbb{N} \
0, text{ otherwise}.end{cases}$$
Then choose the correct option
$a)$ both $f$ and g are Riemann-integrable
$b)$ $f$ is Riemann-integrable but $g$ is not
$c)$ $g$ is Riemann-integrable but $f$ is not
$d)$ neither $f$ nor $g$ is Riemann-integrable
i thinks option $d)$ will correct that is neither $f$ nor $g$ is Riemann-integrable because $f$ and $g$ are not continuous.
Is its True ??
real-analysis integration riemann-integration
edited Nov 20 '18 at 23:40
Batominovski
33.9k33292
asked Nov 20 '18 at 23:26
Messi fifa
51611
add a comment |
let $f : [0,1] rightarrow mathbb{R}$ and $g:[0,1] rightarrow mathbb{R}$ be two function define by
$$f(x) =
begin{cases}
frac {1}{n},text{ if }x = frac{1}{n},n in mathbb{N}\
0, text{otherwise}end{cases}$$ and $$g(x) =
begin{cases}
n,text{ if } x = frac{1}{n}, n in mathbb{N} \
0, text{ otherwise}.end{cases}$$
Then choose the correct option
$a)$ both $f$ and g are Riemann-integrable
$b)$ $f$ is Riemann-integrable but $g$ is not
$c)$ $g$ is Riemann-integrable but $f$ is not
$d)$ neither $f$ nor $g$ is Riemann-integrable
i thinks option $d)$ will correct that is neither $f$ nor $g$ is Riemann-integrable because $f$ and $g$ are not continuous.
Is its True ??
real-analysis integration riemann-integration
edited Nov 20 '18 at 23:40
Batominovski
33.9k33292
asked Nov 20 '18 at 23:26
Messi fifa
51611
2
Is it true that Riemann integrable functions must be continuous?
– Saucy O'Path
Nov 20 '18 at 23:28
@SaucyO'Path ya i missed that
– Messi fifa
Nov 20 '18 at 23:35
add a comment |
let $f : [0,1] rightarrow mathbb{R}$ and $g:[0,1] rightarrow mathbb{R}$ be two function define by
$$f(x) =
begin{cases}
frac {1}{n},text{ if }x = frac{1}{n},n in mathbb{N}\
0, text{otherwise}end{cases}$$ and $$g(x) =
begin{cases}
n,text{ if } x = frac{1}{n}, n in mathbb{N} \
0, text{ otherwise}.end{cases}$$
Then choose the correct option
$a)$ both $f$ and g are Riemann-integrable
$b)$ $f$ is Riemann-integrable but $g$ is not
$c)$ $g$ is Riemann-integrable but $f$ is not
$d)$ neither $f$ nor $g$ is Riemann-integrable
i thinks option $d)$ will correct that is neither $f$ nor $g$ is Riemann-integrable because $f$ and $g$ are not continuous.
Is its True ??
real-analysis integration riemann-integration
edited Nov 20 '18 at 23:40
Batominovski
33.9k33292
asked Nov 20 '18 at 23:26
Messi fifa
51611
let $f : [0,1] rightarrow mathbb{R}$ and $g:[0,1] rightarrow mathbb{R}$ be two function define by
$$f(x) =
begin{cases}
frac {1}{n},text{ if }x = frac{1}{n},n in mathbb{N}\
0, text{otherwise}end{cases}$$ and $$g(x) =
begin{cases}
n,text{ if } x = frac{1}{n}, n in mathbb{N} \
0, text{ otherwise}.end{cases}$$
Then choose the correct option
$a)$ both $f$ and g are Riemann-integrable
$b)$ $f$ is Riemann-integrable but $g$ is not
$c)$ $g$ is Riemann-integrable but $f$ is not
$d)$ neither $f$ nor $g$ is Riemann-integrable
i thinks option $d)$ will correct that is neither $f$ nor $g$ is Riemann-integrable because $f$ and $g$ are not continuous.
Is its True ??
real-analysis integration riemann-integration
real-analysis integration riemann-integration
edited Nov 20 '18 at 23:40
Batominovski
33.9k33292
asked Nov 20 '18 at 23:26
Messi fifa
51611
edited Nov 20 '18 at 23:40
Batominovski
33.9k33292
asked Nov 20 '18 at 23:26
Messi fifa
51611
edited Nov 20 '18 at 23:40
Batominovski
33.9k33292
edited Nov 20 '18 at 23:40
Batominovski
33.9k33292
edited Nov 20 '18 at 23:40
Batominovski
33.9k33292
33.9k33292
asked Nov 20 '18 at 23:26
Messi fifa
51611
asked Nov 20 '18 at 23:26
Messi fifa
51611
asked Nov 20 '18 at 23:26
Messi fifa
51611
51611
2
Is it true that Riemann integrable functions must be continuous?
– Saucy O'Path
Nov 20 '18 at 23:28
@SaucyO'Path ya i missed that
– Messi fifa
Nov 20 '18 at 23:35
add a comment |
2
Is it true that Riemann integrable functions must be continuous?
– Saucy O'Path
Nov 20 '18 at 23:28
@SaucyO'Path ya i missed that
– Messi fifa
Nov 20 '18 at 23:35
2
2
Is it true that Riemann integrable functions must be continuous?
– Saucy O'Path
Nov 20 '18 at 23:28
Is it true that Riemann integrable functions must be continuous?
– Saucy O'Path
Nov 20 '18 at 23:28
@SaucyO'Path ya i missed that
– Messi fifa
Nov 20 '18 at 23:35
@SaucyO'Path ya i missed that
– Messi fifa
Nov 20 '18 at 23:35
add a comment |
1 Answer
1
active
oldest
votes
b) is the correct answer. $g$ is not even a bounded function so it is not a Riemann integrable. $f$ is Riemann integrable because it is bounded and continuous almost everywhere.
answered Nov 20 '18 at 23:29
Kavi Rama Murthy
51k31854
one more doubts @Kavi sir $g(x)$ has countable point of discontinuity i mean it can be reimann integrable ??
– Messi fifa
Nov 20 '18 at 23:35
1
Yes, any bounded function with only countable number of discontinuities is Riemann integrable.
– Kavi Rama Murthy
Nov 20 '18 at 23:36
okkss thanks u sir i gots its
– Messi fifa
Nov 20 '18 at 23:37
add a comment |
Your Answer
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1 Answer
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votes
b) is the correct answer. $g$ is not even a bounded function so it is not a Riemann integrable. $f$ is Riemann integrable because it is bounded and continuous almost everywhere.
answered Nov 20 '18 at 23:29
Kavi Rama Murthy
51k31854
one more doubts @Kavi sir $g(x)$ has countable point of discontinuity i mean it can be reimann integrable ??
– Messi fifa
Nov 20 '18 at 23:35
1
Yes, any bounded function with only countable number of discontinuities is Riemann integrable.
– Kavi Rama Murthy
Nov 20 '18 at 23:36
okkss thanks u sir i gots its
– Messi fifa
Nov 20 '18 at 23:37
add a comment |
b) is the correct answer. $g$ is not even a bounded function so it is not a Riemann integrable. $f$ is Riemann integrable because it is bounded and continuous almost everywhere.
answered Nov 20 '18 at 23:29
Kavi Rama Murthy
51k31854
one more doubts @Kavi sir $g(x)$ has countable point of discontinuity i mean it can be reimann integrable ??
– Messi fifa
Nov 20 '18 at 23:35
1
Yes, any bounded function with only countable number of discontinuities is Riemann integrable.
– Kavi Rama Murthy
Nov 20 '18 at 23:36
okkss thanks u sir i gots its
– Messi fifa
Nov 20 '18 at 23:37
add a comment |
b) is the correct answer. $g$ is not even a bounded function so it is not a Riemann integrable. $f$ is Riemann integrable because it is bounded and continuous almost everywhere.
answered Nov 20 '18 at 23:29
Kavi Rama Murthy
51k31854
b) is the correct answer. $g$ is not even a bounded function so it is not a Riemann integrable. $f$ is Riemann integrable because it is bounded and continuous almost everywhere.
answered Nov 20 '18 at 23:29
Kavi Rama Murthy
51k31854
answered Nov 20 '18 at 23:29
Kavi Rama Murthy
51k31854
answered Nov 20 '18 at 23:29
Kavi Rama Murthy
51k31854
answered Nov 20 '18 at 23:29
Kavi Rama Murthy
51k31854
51k31854
one more doubts @Kavi sir $g(x)$ has countable point of discontinuity i mean it can be reimann integrable ??
– Messi fifa
Nov 20 '18 at 23:35
1
Yes, any bounded function with only countable number of discontinuities is Riemann integrable.
– Kavi Rama Murthy
Nov 20 '18 at 23:36
okkss thanks u sir i gots its
– Messi fifa
Nov 20 '18 at 23:37
add a comment |
one more doubts @Kavi sir $g(x)$ has countable point of discontinuity i mean it can be reimann integrable ??
– Messi fifa
Nov 20 '18 at 23:35
1
Yes, any bounded function with only countable number of discontinuities is Riemann integrable.
– Kavi Rama Murthy
Nov 20 '18 at 23:36
okkss thanks u sir i gots its
– Messi fifa
Nov 20 '18 at 23:37
one more doubts @Kavi sir $g(x)$ has countable point of discontinuity i mean it can be reimann integrable ??
– Messi fifa
Nov 20 '18 at 23:35
one more doubts @Kavi sir $g(x)$ has countable point of discontinuity i mean it can be reimann integrable ??
– Messi fifa
Nov 20 '18 at 23:35
1
1
Yes, any bounded function with only countable number of discontinuities is Riemann integrable.
– Kavi Rama Murthy
Nov 20 '18 at 23:36
Yes, any bounded function with only countable number of discontinuities is Riemann integrable.
– Kavi Rama Murthy
Nov 20 '18 at 23:36
okkss thanks u sir i gots its
– Messi fifa
Nov 20 '18 at 23:37
okkss thanks u sir i gots its
– Messi fifa
Nov 20 '18 at 23:37
add a comment |
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2
Is it true that Riemann integrable functions must be continuous?
– Saucy O'Path
Nov 20 '18 at 23:28
@SaucyO'Path ya i missed that
– Messi fifa
Nov 20 '18 at 23:35