Mixing
$L^1$ interesting property
$begingroup$
Is it true that if $f:[0,1]tomathbb{R}$ is in $L^1 (0,1)$ and
$$int_0^t f(s) ds = 0, forall tin [0,1]$$, then $f(t) =0$ almost everywhere?
measure-theory lebesgue-measure
edited Jan 27 at 18:01
Anubhab Ghosal
1,20619
asked Jan 27 at 17:27
BogdanBogdan
67349
$endgroup$
add a comment |
$begingroup$
Is it true that if $f:[0,1]tomathbb{R}$ is in $L^1 (0,1)$ and
$$int_0^t f(s) ds = 0, forall tin [0,1]$$, then $f(t) =0$ almost everywhere?
measure-theory lebesgue-measure
edited Jan 27 at 18:01
Anubhab Ghosal
1,20619
asked Jan 27 at 17:27
BogdanBogdan
67349
$endgroup$
$begingroup$
Yes, it certainly is true.
$endgroup$
– Umberto P.
Jan 27 at 17:29
$begingroup$
Why is it true? Can you give some details. We do not know that f is positive.
$endgroup$
– Bogdan
Jan 27 at 17:32
add a comment |
$begingroup$
Is it true that if $f:[0,1]tomathbb{R}$ is in $L^1 (0,1)$ and
$$int_0^t f(s) ds = 0, forall tin [0,1]$$, then $f(t) =0$ almost everywhere?
measure-theory lebesgue-measure
edited Jan 27 at 18:01
Anubhab Ghosal
1,20619
asked Jan 27 at 17:27
BogdanBogdan
67349
$endgroup$
Is it true that if $f:[0,1]tomathbb{R}$ is in $L^1 (0,1)$ and
$$int_0^t f(s) ds = 0, forall tin [0,1]$$, then $f(t) =0$ almost everywhere?
measure-theory lebesgue-measure
measure-theory lebesgue-measure
edited Jan 27 at 18:01
Anubhab Ghosal
1,20619
asked Jan 27 at 17:27
BogdanBogdan
67349
edited Jan 27 at 18:01
Anubhab Ghosal
1,20619
asked Jan 27 at 17:27
BogdanBogdan
67349
edited Jan 27 at 18:01
Anubhab Ghosal
1,20619
edited Jan 27 at 18:01
Anubhab Ghosal
1,20619
edited Jan 27 at 18:01
Anubhab Ghosal
1,20619
1,20619
asked Jan 27 at 17:27
BogdanBogdan
67349
asked Jan 27 at 17:27
BogdanBogdan
67349
asked Jan 27 at 17:27
BogdanBogdan
67349
67349
$begingroup$
Yes, it certainly is true.
$endgroup$
– Umberto P.
Jan 27 at 17:29
$begingroup$
Why is it true? Can you give some details. We do not know that f is positive.
$endgroup$
– Bogdan
Jan 27 at 17:32
add a comment |
$begingroup$
Yes, it certainly is true.
$endgroup$
– Umberto P.
Jan 27 at 17:29
$begingroup$
Why is it true? Can you give some details. We do not know that f is positive.
$endgroup$
– Bogdan
Jan 27 at 17:32
$begingroup$
Yes, it certainly is true.
$endgroup$
– Umberto P.
Jan 27 at 17:29
$begingroup$
Yes, it certainly is true.
$endgroup$
– Umberto P.
Jan 27 at 17:29
$begingroup$
Why is it true? Can you give some details. We do not know that f is positive.
$endgroup$
– Bogdan
Jan 27 at 17:32
$begingroup$
Why is it true? Can you give some details. We do not know that f is positive.
$endgroup$
– Bogdan
Jan 27 at 17:32
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You can differentiate the integral with respect to $t$ almost everywhere, which would imply that $f = 0 $ a.e. A quick way to see this is via Lebesgue points. If $xin (0,1) $ is a Lebesgue point of $f$, then
$$
limlimits_{hto 0} frac{1}{2h}intlimits_{x - h}^{x + h} f(t) dt = f(x).
$$
The integral average equals $0$ since it is the difference of integrals over $[0,x+ h]$ and $[0, x- h]$ and both equal $0$. Hence $f(x) = 0 $ whenever $x$ is a Lebesuge point of $f$. Now, if $fin L^1(0,1)$ almost all points $xin(0,1)$ are Lebesgue points of $f$, hence the claim.
answered Jan 27 at 17:45
HaykHayk
2,6721214
$endgroup$
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can differentiate the integral with respect to $t$ almost everywhere, which would imply that $f = 0 $ a.e. A quick way to see this is via Lebesgue points. If $xin (0,1) $ is a Lebesgue point of $f$, then
$$
limlimits_{hto 0} frac{1}{2h}intlimits_{x - h}^{x + h} f(t) dt = f(x).
$$
The integral average equals $0$ since it is the difference of integrals over $[0,x+ h]$ and $[0, x- h]$ and both equal $0$. Hence $f(x) = 0 $ whenever $x$ is a Lebesuge point of $f$. Now, if $fin L^1(0,1)$ almost all points $xin(0,1)$ are Lebesgue points of $f$, hence the claim.
answered Jan 27 at 17:45
HaykHayk
2,6721214
$endgroup$
add a comment |
$begingroup$
You can differentiate the integral with respect to $t$ almost everywhere, which would imply that $f = 0 $ a.e. A quick way to see this is via Lebesgue points. If $xin (0,1) $ is a Lebesgue point of $f$, then
$$
limlimits_{hto 0} frac{1}{2h}intlimits_{x - h}^{x + h} f(t) dt = f(x).
$$
The integral average equals $0$ since it is the difference of integrals over $[0,x+ h]$ and $[0, x- h]$ and both equal $0$. Hence $f(x) = 0 $ whenever $x$ is a Lebesuge point of $f$. Now, if $fin L^1(0,1)$ almost all points $xin(0,1)$ are Lebesgue points of $f$, hence the claim.
answered Jan 27 at 17:45
HaykHayk
2,6721214
$endgroup$
add a comment |
$begingroup$
You can differentiate the integral with respect to $t$ almost everywhere, which would imply that $f = 0 $ a.e. A quick way to see this is via Lebesgue points. If $xin (0,1) $ is a Lebesgue point of $f$, then
$$
limlimits_{hto 0} frac{1}{2h}intlimits_{x - h}^{x + h} f(t) dt = f(x).
$$
The integral average equals $0$ since it is the difference of integrals over $[0,x+ h]$ and $[0, x- h]$ and both equal $0$. Hence $f(x) = 0 $ whenever $x$ is a Lebesuge point of $f$. Now, if $fin L^1(0,1)$ almost all points $xin(0,1)$ are Lebesgue points of $f$, hence the claim.
answered Jan 27 at 17:45
HaykHayk
2,6721214
$endgroup$
You can differentiate the integral with respect to $t$ almost everywhere, which would imply that $f = 0 $ a.e. A quick way to see this is via Lebesgue points. If $xin (0,1) $ is a Lebesgue point of $f$, then
$$
limlimits_{hto 0} frac{1}{2h}intlimits_{x - h}^{x + h} f(t) dt = f(x).
$$
The integral average equals $0$ since it is the difference of integrals over $[0,x+ h]$ and $[0, x- h]$ and both equal $0$. Hence $f(x) = 0 $ whenever $x$ is a Lebesuge point of $f$. Now, if $fin L^1(0,1)$ almost all points $xin(0,1)$ are Lebesgue points of $f$, hence the claim.
answered Jan 27 at 17:45
HaykHayk
2,6721214
answered Jan 27 at 17:45
HaykHayk
2,6721214
answered Jan 27 at 17:45
HaykHayk
2,6721214
answered Jan 27 at 17:45
HaykHayk
2,6721214
2,6721214
add a comment |
add a comment |
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$begingroup$
Yes, it certainly is true.
$endgroup$
– Umberto P.
Jan 27 at 17:29
$begingroup$
Why is it true? Can you give some details. We do not know that f is positive.
$endgroup$
– Bogdan
Jan 27 at 17:32