Show that $f$ is Lebesgue integrable
0
$begingroup$
I am given a function defined on $(mathbb R,$ Borelians, Lebesgue measure)
such that
$$f= begin{cases} + infty &text{if } x=0 \
ln(|x|) &text{if } 0lt |x|lt1 \
0 &text{if } |x|ge1 end{cases}
$$
How can I show that this function is Lebesgue-integrable ?
real-analysis measure-theory
edited Dec 1 '18 at 14:51
Daniele Tampieri
1,9561619
asked Dec 1 '18 at 14:34
Jonathan BaramJonathan Baram
140113
$endgroup$
add a comment |
0
$begingroup$
I am given a function defined on $(mathbb R,$ Borelians, Lebesgue measure)
such that
$$f= begin{cases} + infty &text{if } x=0 \
ln(|x|) &text{if } 0lt |x|lt1 \
0 &text{if } |x|ge1 end{cases}
$$
How can I show that this function is Lebesgue-integrable ?
real-analysis measure-theory
edited Dec 1 '18 at 14:51
Daniele Tampieri
1,9561619
asked Dec 1 '18 at 14:34
Jonathan BaramJonathan Baram
140113
$endgroup$
$begingroup$
Maybe you can use that the riemann integral of $f$ on $[-1, - 1/n] cup [1/n, 1]$ for $ngeq 2$ is equal to the Lebesgue integral on the same interval. Construct a sequence of functions based on the previous observation and then try to apply monotone convergence theorem somewhere. What did you try btw? Do you know that the riemann integral is equal to the lebesgue integral?
$endgroup$
– Shashi
Dec 1 '18 at 16:15
add a comment |
0
0
0
$begingroup$
I am given a function defined on $(mathbb R,$ Borelians, Lebesgue measure)
such that
$$f= begin{cases} + infty &text{if } x=0 \
ln(|x|) &text{if } 0lt |x|lt1 \
0 &text{if } |x|ge1 end{cases}
$$
How can I show that this function is Lebesgue-integrable ?
real-analysis measure-theory
edited Dec 1 '18 at 14:51
Daniele Tampieri
1,9561619
asked Dec 1 '18 at 14:34
Jonathan BaramJonathan Baram
140113
$endgroup$
I am given a function defined on $(mathbb R,$ Borelians, Lebesgue measure)
such that
$$f= begin{cases} + infty &text{if } x=0 \
ln(|x|) &text{if } 0lt |x|lt1 \
0 &text{if } |x|ge1 end{cases}
$$
How can I show that this function is Lebesgue-integrable ?
real-analysis measure-theory
real-analysis measure-theory
edited Dec 1 '18 at 14:51
Daniele Tampieri
1,9561619
asked Dec 1 '18 at 14:34
Jonathan BaramJonathan Baram
140113
edited Dec 1 '18 at 14:51
Daniele Tampieri
1,9561619
asked Dec 1 '18 at 14:34
Jonathan BaramJonathan Baram
140113
edited Dec 1 '18 at 14:51
Daniele Tampieri
1,9561619
edited Dec 1 '18 at 14:51
Daniele Tampieri
1,9561619
edited Dec 1 '18 at 14:51
Daniele Tampieri
1,9561619
1,9561619
asked Dec 1 '18 at 14:34
Jonathan BaramJonathan Baram
140113
asked Dec 1 '18 at 14:34
Jonathan BaramJonathan Baram
140113
asked Dec 1 '18 at 14:34
Jonathan BaramJonathan Baram
140113
140113
$begingroup$
Maybe you can use that the riemann integral of $f$ on $[-1, - 1/n] cup [1/n, 1]$ for $ngeq 2$ is equal to the Lebesgue integral on the same interval. Construct a sequence of functions based on the previous observation and then try to apply monotone convergence theorem somewhere. What did you try btw? Do you know that the riemann integral is equal to the lebesgue integral?
$endgroup$
– Shashi
Dec 1 '18 at 16:15
add a comment |
$begingroup$
Maybe you can use that the riemann integral of $f$ on $[-1, - 1/n] cup [1/n, 1]$ for $ngeq 2$ is equal to the Lebesgue integral on the same interval. Construct a sequence of functions based on the previous observation and then try to apply monotone convergence theorem somewhere. What did you try btw? Do you know that the riemann integral is equal to the lebesgue integral?
$endgroup$
– Shashi
Dec 1 '18 at 16:15
$begingroup$
Maybe you can use that the riemann integral of $f$ on $[-1, - 1/n] cup [1/n, 1]$ for $ngeq 2$ is equal to the Lebesgue integral on the same interval. Construct a sequence of functions based on the previous observation and then try to apply monotone convergence theorem somewhere. What did you try btw? Do you know that the riemann integral is equal to the lebesgue integral?
$endgroup$
– Shashi
Dec 1 '18 at 16:15
$begingroup$
Maybe you can use that the riemann integral of $f$ on $[-1, - 1/n] cup [1/n, 1]$ for $ngeq 2$ is equal to the Lebesgue integral on the same interval. Construct a sequence of functions based on the previous observation and then try to apply monotone convergence theorem somewhere. What did you try btw? Do you know that the riemann integral is equal to the lebesgue integral?
$endgroup$
– Shashi
Dec 1 '18 at 16:15
add a comment |
1 Answer
1
active
oldest
votes
-1
$begingroup$
Since $f$ is even it is enough to show that it is integrable on $(0,1)$. But $$int_{(0,1)} f(u)du =ulnu-u|_0^1 =1$$
answered Dec 1 '18 at 15:01
MotylaNogaTomkaMazuraMotylaNogaTomkaMazura
6,542917
$endgroup$
$begingroup$
isn't there a way to show it using the definition of Lebesgue-integrability and basic theorems such as dominated convergence, monotone convergence?
$endgroup$
– Jonathan Baram
Dec 1 '18 at 15:12
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
-1
$begingroup$
Since $f$ is even it is enough to show that it is integrable on $(0,1)$. But $$int_{(0,1)} f(u)du =ulnu-u|_0^1 =1$$
answered Dec 1 '18 at 15:01
MotylaNogaTomkaMazuraMotylaNogaTomkaMazura
6,542917
$endgroup$
$begingroup$
isn't there a way to show it using the definition of Lebesgue-integrability and basic theorems such as dominated convergence, monotone convergence?
$endgroup$
– Jonathan Baram
Dec 1 '18 at 15:12
add a comment |
-1
$begingroup$
Since $f$ is even it is enough to show that it is integrable on $(0,1)$. But $$int_{(0,1)} f(u)du =ulnu-u|_0^1 =1$$
answered Dec 1 '18 at 15:01
MotylaNogaTomkaMazuraMotylaNogaTomkaMazura
6,542917
$endgroup$
$begingroup$
isn't there a way to show it using the definition of Lebesgue-integrability and basic theorems such as dominated convergence, monotone convergence?
$endgroup$
– Jonathan Baram
Dec 1 '18 at 15:12
add a comment |
-1
-1
-1
$begingroup$
Since $f$ is even it is enough to show that it is integrable on $(0,1)$. But $$int_{(0,1)} f(u)du =ulnu-u|_0^1 =1$$
answered Dec 1 '18 at 15:01
MotylaNogaTomkaMazuraMotylaNogaTomkaMazura
6,542917
$endgroup$
Since $f$ is even it is enough to show that it is integrable on $(0,1)$. But $$int_{(0,1)} f(u)du =ulnu-u|_0^1 =1$$
answered Dec 1 '18 at 15:01
MotylaNogaTomkaMazuraMotylaNogaTomkaMazura
6,542917
answered Dec 1 '18 at 15:01
MotylaNogaTomkaMazuraMotylaNogaTomkaMazura
6,542917
answered Dec 1 '18 at 15:01
MotylaNogaTomkaMazuraMotylaNogaTomkaMazura
6,542917
answered Dec 1 '18 at 15:01
MotylaNogaTomkaMazuraMotylaNogaTomkaMazura
6,542917
6,542917
$begingroup$
isn't there a way to show it using the definition of Lebesgue-integrability and basic theorems such as dominated convergence, monotone convergence?
$endgroup$
– Jonathan Baram
Dec 1 '18 at 15:12
add a comment |
$begingroup$
isn't there a way to show it using the definition of Lebesgue-integrability and basic theorems such as dominated convergence, monotone convergence?
$endgroup$
– Jonathan Baram
Dec 1 '18 at 15:12
$begingroup$
isn't there a way to show it using the definition of Lebesgue-integrability and basic theorems such as dominated convergence, monotone convergence?
$endgroup$
– Jonathan Baram
Dec 1 '18 at 15:12
$begingroup$
isn't there a way to show it using the definition of Lebesgue-integrability and basic theorems such as dominated convergence, monotone convergence?
$endgroup$
– Jonathan Baram
Dec 1 '18 at 15:12
add a comment |
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$begingroup$
Maybe you can use that the riemann integral of $f$ on $[-1, - 1/n] cup [1/n, 1]$ for $ngeq 2$ is equal to the Lebesgue integral on the same interval. Construct a sequence of functions based on the previous observation and then try to apply monotone convergence theorem somewhere. What did you try btw? Do you know that the riemann integral is equal to the lebesgue integral?
$endgroup$
– Shashi
Dec 1 '18 at 16:15