Showing $f(x) = frac{1}{sqrt x}$ is Lebesgue integrable on $(0,1]$
$begingroup$
While reviewing for an exam recently, I came across this question which gave me pause.
Explain why $f(x) = frac{1}{sqrt x}$ is Lebesgue integrable over
$(0,1]$.
It is clear that $f$ is a decreasing, non-negative function. So, it is called Lebesgue integrable over $(0,1]$ if:
$int_{(0,1]} f text{dm} < infty$
And:
$int_{(0,1]} f text{dm} = sup{ int_{(0,1]} s text{dm}: s text{ simple}, s(x) leq f(x) forall x in (0,1]}$
The problem I’m having is that a lot of the typical useful theorems (MCT, DCT) are about non-decreasing functions.
Broadly speaking, I understand that the “problem” with this function is that $f(0)$ is undefined and as $x to 0$, $f(x)$ gets very big.
I guess that the reason this function is integrable over $(0,1]$ is that the most “problematic” point ($x = 0$) is removed.
How do I go about showing this more rigorously? Which (common) theorems should I use?
real-analysis integration analysis lebesgue-integral
edited Jan 14 at 22:48
Neil hawking
1
asked Dec 12 '18 at 19:46
TuringTester69TuringTester69
307213
$endgroup$
add a comment |
$begingroup$
While reviewing for an exam recently, I came across this question which gave me pause.
Explain why $f(x) = frac{1}{sqrt x}$ is Lebesgue integrable over
$(0,1]$.
It is clear that $f$ is a decreasing, non-negative function. So, it is called Lebesgue integrable over $(0,1]$ if:
$int_{(0,1]} f text{dm} < infty$
And:
$int_{(0,1]} f text{dm} = sup{ int_{(0,1]} s text{dm}: s text{ simple}, s(x) leq f(x) forall x in (0,1]}$
The problem I’m having is that a lot of the typical useful theorems (MCT, DCT) are about non-decreasing functions.
Broadly speaking, I understand that the “problem” with this function is that $f(0)$ is undefined and as $x to 0$, $f(x)$ gets very big.
I guess that the reason this function is integrable over $(0,1]$ is that the most “problematic” point ($x = 0$) is removed.
How do I go about showing this more rigorously? Which (common) theorems should I use?
real-analysis integration analysis lebesgue-integral
edited Jan 14 at 22:48
Neil hawking
1
asked Dec 12 '18 at 19:46
TuringTester69TuringTester69
307213
$endgroup$
$begingroup$
Could you, instead, think about integration of $g(x)=dfrac{1}{sqrt{1-x}}$ on $[0,1)?$ And then argue that $displaystyleint_0^1f(x),dm=int_0^1g(x),dm?$
$endgroup$
– Adrian Keister
Dec 12 '18 at 19:56
add a comment |
$begingroup$
While reviewing for an exam recently, I came across this question which gave me pause.
Explain why $f(x) = frac{1}{sqrt x}$ is Lebesgue integrable over
$(0,1]$.
It is clear that $f$ is a decreasing, non-negative function. So, it is called Lebesgue integrable over $(0,1]$ if:
$int_{(0,1]} f text{dm} < infty$
And:
$int_{(0,1]} f text{dm} = sup{ int_{(0,1]} s text{dm}: s text{ simple}, s(x) leq f(x) forall x in (0,1]}$
The problem I’m having is that a lot of the typical useful theorems (MCT, DCT) are about non-decreasing functions.
Broadly speaking, I understand that the “problem” with this function is that $f(0)$ is undefined and as $x to 0$, $f(x)$ gets very big.
I guess that the reason this function is integrable over $(0,1]$ is that the most “problematic” point ($x = 0$) is removed.
How do I go about showing this more rigorously? Which (common) theorems should I use?
real-analysis integration analysis lebesgue-integral
edited Jan 14 at 22:48
Neil hawking
1
asked Dec 12 '18 at 19:46
TuringTester69TuringTester69
307213
$endgroup$
While reviewing for an exam recently, I came across this question which gave me pause.
Explain why $f(x) = frac{1}{sqrt x}$ is Lebesgue integrable over
$(0,1]$.
It is clear that $f$ is a decreasing, non-negative function. So, it is called Lebesgue integrable over $(0,1]$ if:
$int_{(0,1]} f text{dm} < infty$
And:
$int_{(0,1]} f text{dm} = sup{ int_{(0,1]} s text{dm}: s text{ simple}, s(x) leq f(x) forall x in (0,1]}$
The problem I’m having is that a lot of the typical useful theorems (MCT, DCT) are about non-decreasing functions.
Broadly speaking, I understand that the “problem” with this function is that $f(0)$ is undefined and as $x to 0$, $f(x)$ gets very big.
I guess that the reason this function is integrable over $(0,1]$ is that the most “problematic” point ($x = 0$) is removed.
How do I go about showing this more rigorously? Which (common) theorems should I use?
real-analysis integration analysis lebesgue-integral
real-analysis integration analysis lebesgue-integral
edited Jan 14 at 22:48
Neil hawking
1
asked Dec 12 '18 at 19:46
TuringTester69TuringTester69
307213
edited Jan 14 at 22:48
Neil hawking
1
asked Dec 12 '18 at 19:46
TuringTester69TuringTester69
307213
edited Jan 14 at 22:48
Neil hawking
1
edited Jan 14 at 22:48
Neil hawking
1
edited Jan 14 at 22:48
Neil hawking
1
1
asked Dec 12 '18 at 19:46
TuringTester69TuringTester69
307213
asked Dec 12 '18 at 19:46
TuringTester69TuringTester69
307213
asked Dec 12 '18 at 19:46
TuringTester69TuringTester69
307213
307213
$begingroup$
Could you, instead, think about integration of $g(x)=dfrac{1}{sqrt{1-x}}$ on $[0,1)?$ And then argue that $displaystyleint_0^1f(x),dm=int_0^1g(x),dm?$
$endgroup$
– Adrian Keister
Dec 12 '18 at 19:56
add a comment |
$begingroup$
Could you, instead, think about integration of $g(x)=dfrac{1}{sqrt{1-x}}$ on $[0,1)?$ And then argue that $displaystyleint_0^1f(x),dm=int_0^1g(x),dm?$
$endgroup$
– Adrian Keister
Dec 12 '18 at 19:56
$begingroup$
Could you, instead, think about integration of $g(x)=dfrac{1}{sqrt{1-x}}$ on $[0,1)?$ And then argue that $displaystyleint_0^1f(x),dm=int_0^1g(x),dm?$
$endgroup$
– Adrian Keister
Dec 12 '18 at 19:56
$begingroup$
Could you, instead, think about integration of $g(x)=dfrac{1}{sqrt{1-x}}$ on $[0,1)?$ And then argue that $displaystyleint_0^1f(x),dm=int_0^1g(x),dm?$
$endgroup$
– Adrian Keister
Dec 12 '18 at 19:56
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$int f_{(0,1]}=int lim fchi_{[1/n,1]}=limint fchi_{[1/n,1]}=lim (2-sqrt {2/n)}=2$,
the first equality is obvious, the second MCT, and the third is true because the Riemann integral coincides with the Lebesgue integral on $[1/n,1].$
As you point out, one strategy for doing these problems is to "back off" the bad point and use MCT or DCT.
answered Dec 12 '18 at 20:04
MatematletaMatematleta
11.4k2920
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add a comment |
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1 Answer
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$begingroup$
$int f_{(0,1]}=int lim fchi_{[1/n,1]}=limint fchi_{[1/n,1]}=lim (2-sqrt {2/n)}=2$,
the first equality is obvious, the second MCT, and the third is true because the Riemann integral coincides with the Lebesgue integral on $[1/n,1].$
As you point out, one strategy for doing these problems is to "back off" the bad point and use MCT or DCT.
answered Dec 12 '18 at 20:04
MatematletaMatematleta
11.4k2920
$endgroup$
add a comment |
$begingroup$
$int f_{(0,1]}=int lim fchi_{[1/n,1]}=limint fchi_{[1/n,1]}=lim (2-sqrt {2/n)}=2$,
the first equality is obvious, the second MCT, and the third is true because the Riemann integral coincides with the Lebesgue integral on $[1/n,1].$
As you point out, one strategy for doing these problems is to "back off" the bad point and use MCT or DCT.
answered Dec 12 '18 at 20:04
MatematletaMatematleta
11.4k2920
$endgroup$
add a comment |
$begingroup$
$int f_{(0,1]}=int lim fchi_{[1/n,1]}=limint fchi_{[1/n,1]}=lim (2-sqrt {2/n)}=2$,
the first equality is obvious, the second MCT, and the third is true because the Riemann integral coincides with the Lebesgue integral on $[1/n,1].$
As you point out, one strategy for doing these problems is to "back off" the bad point and use MCT or DCT.
answered Dec 12 '18 at 20:04
MatematletaMatematleta
11.4k2920
$endgroup$
$int f_{(0,1]}=int lim fchi_{[1/n,1]}=limint fchi_{[1/n,1]}=lim (2-sqrt {2/n)}=2$,
the first equality is obvious, the second MCT, and the third is true because the Riemann integral coincides with the Lebesgue integral on $[1/n,1].$
As you point out, one strategy for doing these problems is to "back off" the bad point and use MCT or DCT.
answered Dec 12 '18 at 20:04
MatematletaMatematleta
11.4k2920
answered Dec 12 '18 at 20:04
MatematletaMatematleta
11.4k2920
answered Dec 12 '18 at 20:04
MatematletaMatematleta
11.4k2920
answered Dec 12 '18 at 20:04
MatematletaMatematleta
11.4k2920
11.4k2920
add a comment |
add a comment |
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$begingroup$
Could you, instead, think about integration of $g(x)=dfrac{1}{sqrt{1-x}}$ on $[0,1)?$ And then argue that $displaystyleint_0^1f(x),dm=int_0^1g(x),dm?$
$endgroup$
– Adrian Keister
Dec 12 '18 at 19:56