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Any hypotheses missing in this measure-theoretic version of Leibniz’s integral rule?
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I have seen the following measure-theoretic version of Leibniz’s integral rule:
Let $a,b in mathbbR,$ $a<b,$ $I$ an open interval, and let $f colon [a,b] times I to mathbbR.$ Assume that, for each $x in [a,b],$ $f(x,cdot) colon I to mathbbR$ is differentiable at every $t in I.$ Also assume that the function beginequation* F(t)= int_a^b f(x,t)dx endequation* is well defined for $t in I.$ Under these conditions, given $t_0 in I,$ if there exist an integrable function $phi colon [a,b] to mathbbR$ and a $delta > 0$ such that $|f(x,cdot)'(t)| leq phi(x)$ for all $x in [a,b], t in (t_0 -delta, t_0 + delta) subseteq I,$ then $F$ is differentiable at $t_0$ and beginequation* F'(t_0) = int_a ^b f(x,cdot)'(t_0)dx.endequation*
I would like to know whether the hypotheses in the statement imply that $x mapsto f(x,cdot)'(t_0)$ is integrable or this needs to be added as an extra condition.
Thank you four your help.
real-analysis measure-theory
asked Aug 2 at 14:25
VerÃsimo
14
 |Â
show 5 more comments
up vote
-1
down vote
favorite
I have seen the following measure-theoretic version of Leibniz’s integral rule:
Let $a,b in mathbbR,$ $a<b,$ $I$ an open interval, and let $f colon [a,b] times I to mathbbR.$ Assume that, for each $x in [a,b],$ $f(x,cdot) colon I to mathbbR$ is differentiable at every $t in I.$ Also assume that the function beginequation* F(t)= int_a^b f(x,t)dx endequation* is well defined for $t in I.$ Under these conditions, given $t_0 in I,$ if there exist an integrable function $phi colon [a,b] to mathbbR$ and a $delta > 0$ such that $|f(x,cdot)'(t)| leq phi(x)$ for all $x in [a,b], t in (t_0 -delta, t_0 + delta) subseteq I,$ then $F$ is differentiable at $t_0$ and beginequation* F'(t_0) = int_a ^b f(x,cdot)'(t_0)dx.endequation*
I would like to know whether the hypotheses in the statement imply that $x mapsto f(x,cdot)'(t_0)$ is integrable or this needs to be added as an extra condition.
Thank you four your help.
real-analysis measure-theory
asked Aug 2 at 14:25
VerÃsimo
14
What do you think is missing?
– uniquesolution
Aug 2 at 14:31
Sorry, I unintentionally posted the question before I finished writing it. Now it is complete.
– VerÃsimo
Aug 2 at 14:34
Yes, it is integrable, because its absolute value is bounded by an integrable function.
– uniquesolution
Aug 2 at 14:36
That argument works if we already know that the function is measurable. Do we know that?
– VerÃsimo
Aug 2 at 14:37
1
It represents the variable of the function. $f(x, cdot)$ denotes, for a fixed $x,$ the function that takes each $t in I$ to $f(x,t).$
– VerÃsimo
Aug 2 at 15:29
 |Â
show 5 more comments
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
I have seen the following measure-theoretic version of Leibniz’s integral rule:
Let $a,b in mathbbR,$ $a<b,$ $I$ an open interval, and let $f colon [a,b] times I to mathbbR.$ Assume that, for each $x in [a,b],$ $f(x,cdot) colon I to mathbbR$ is differentiable at every $t in I.$ Also assume that the function beginequation* F(t)= int_a^b f(x,t)dx endequation* is well defined for $t in I.$ Under these conditions, given $t_0 in I,$ if there exist an integrable function $phi colon [a,b] to mathbbR$ and a $delta > 0$ such that $|f(x,cdot)'(t)| leq phi(x)$ for all $x in [a,b], t in (t_0 -delta, t_0 + delta) subseteq I,$ then $F$ is differentiable at $t_0$ and beginequation* F'(t_0) = int_a ^b f(x,cdot)'(t_0)dx.endequation*
I would like to know whether the hypotheses in the statement imply that $x mapsto f(x,cdot)'(t_0)$ is integrable or this needs to be added as an extra condition.
Thank you four your help.
real-analysis measure-theory
asked Aug 2 at 14:25
VerÃsimo
14
I have seen the following measure-theoretic version of Leibniz’s integral rule:
Let $a,b in mathbbR,$ $a<b,$ $I$ an open interval, and let $f colon [a,b] times I to mathbbR.$ Assume that, for each $x in [a,b],$ $f(x,cdot) colon I to mathbbR$ is differentiable at every $t in I.$ Also assume that the function beginequation* F(t)= int_a^b f(x,t)dx endequation* is well defined for $t in I.$ Under these conditions, given $t_0 in I,$ if there exist an integrable function $phi colon [a,b] to mathbbR$ and a $delta > 0$ such that $|f(x,cdot)'(t)| leq phi(x)$ for all $x in [a,b], t in (t_0 -delta, t_0 + delta) subseteq I,$ then $F$ is differentiable at $t_0$ and beginequation* F'(t_0) = int_a ^b f(x,cdot)'(t_0)dx.endequation*
I would like to know whether the hypotheses in the statement imply that $x mapsto f(x,cdot)'(t_0)$ is integrable or this needs to be added as an extra condition.
Thank you four your help.
real-analysis measure-theory
asked Aug 2 at 14:25
VerÃsimo
14
edited Aug 2 at 14:33
asked Aug 2 at 14:25
VerÃsimo
14
asked Aug 2 at 14:25
VerÃsimo
14
asked Aug 2 at 14:25
VerÃsimo
14
14
What do you think is missing?
– uniquesolution
Aug 2 at 14:31
Sorry, I unintentionally posted the question before I finished writing it. Now it is complete.
– VerÃsimo
Aug 2 at 14:34
Yes, it is integrable, because its absolute value is bounded by an integrable function.
– uniquesolution
Aug 2 at 14:36
That argument works if we already know that the function is measurable. Do we know that?
– VerÃsimo
Aug 2 at 14:37
1
It represents the variable of the function. $f(x, cdot)$ denotes, for a fixed $x,$ the function that takes each $t in I$ to $f(x,t).$
– VerÃsimo
Aug 2 at 15:29
 |Â
show 5 more comments
What do you think is missing?
– uniquesolution
Aug 2 at 14:31
Sorry, I unintentionally posted the question before I finished writing it. Now it is complete.
– VerÃsimo
Aug 2 at 14:34
Yes, it is integrable, because its absolute value is bounded by an integrable function.
– uniquesolution
Aug 2 at 14:36
That argument works if we already know that the function is measurable. Do we know that?
– VerÃsimo
Aug 2 at 14:37
1
It represents the variable of the function. $f(x, cdot)$ denotes, for a fixed $x,$ the function that takes each $t in I$ to $f(x,t).$
– VerÃsimo
Aug 2 at 15:29
What do you think is missing?
– uniquesolution
Aug 2 at 14:31
What do you think is missing?
– uniquesolution
Aug 2 at 14:31
Sorry, I unintentionally posted the question before I finished writing it. Now it is complete.
– VerÃsimo
Aug 2 at 14:34
Sorry, I unintentionally posted the question before I finished writing it. Now it is complete.
– VerÃsimo
Aug 2 at 14:34
Yes, it is integrable, because its absolute value is bounded by an integrable function.
– uniquesolution
Aug 2 at 14:36
Yes, it is integrable, because its absolute value is bounded by an integrable function.
– uniquesolution
Aug 2 at 14:36
That argument works if we already know that the function is measurable. Do we know that?
– VerÃsimo
Aug 2 at 14:37
That argument works if we already know that the function is measurable. Do we know that?
– VerÃsimo
Aug 2 at 14:37
1
1
It represents the variable of the function. $f(x, cdot)$ denotes, for a fixed $x,$ the function that takes each $t in I$ to $f(x,t).$
– VerÃsimo
Aug 2 at 15:29
It represents the variable of the function. $f(x, cdot)$ denotes, for a fixed $x,$ the function that takes each $t in I$ to $f(x,t).$
– VerÃsimo
Aug 2 at 15:29
 |Â
show 5 more comments
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What do you think is missing?
– uniquesolution
Aug 2 at 14:31
Sorry, I unintentionally posted the question before I finished writing it. Now it is complete.
– VerÃsimo
Aug 2 at 14:34
Yes, it is integrable, because its absolute value is bounded by an integrable function.
– uniquesolution
Aug 2 at 14:36
That argument works if we already know that the function is measurable. Do we know that?
– VerÃsimo
Aug 2 at 14:37
1
It represents the variable of the function. $f(x, cdot)$ denotes, for a fixed $x,$ the function that takes each $t in I$ to $f(x,t).$
– VerÃsimo
Aug 2 at 15:29