Show that $f$ is measurable on $(a,b)$
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Let $f,F : (a,b)to mathbb{R}$ such that $F$ is differentiable on $(a,b)$ and $F'(x) =f(x)$ for each $x in (a,b)$. Show that $f$ is measurable on $(a,b)$.
Hint : Show first that $forall x in (a,b) : f(x) = limlimits_{n to infty} n[F(x+frac{1}{n}) - F(x)]$.
I have already proved the hint ... How to proceed please?
measurable-functions
asked Nov 23 at 22:01
Ahmed
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Let $f,F : (a,b)to mathbb{R}$ such that $F$ is differentiable on $(a,b)$ and $F'(x) =f(x)$ for each $x in (a,b)$. Show that $f$ is measurable on $(a,b)$.
Hint : Show first that $forall x in (a,b) : f(x) = limlimits_{n to infty} n[F(x+frac{1}{n}) - F(x)]$.
I have already proved the hint ... How to proceed please?
measurable-functions
asked Nov 23 at 22:01
Ahmed
1,244511
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $f,F : (a,b)to mathbb{R}$ such that $F$ is differentiable on $(a,b)$ and $F'(x) =f(x)$ for each $x in (a,b)$. Show that $f$ is measurable on $(a,b)$.
Hint : Show first that $forall x in (a,b) : f(x) = limlimits_{n to infty} n[F(x+frac{1}{n}) - F(x)]$.
I have already proved the hint ... How to proceed please?
measurable-functions
asked Nov 23 at 22:01
Ahmed
1,244511
Let $f,F : (a,b)to mathbb{R}$ such that $F$ is differentiable on $(a,b)$ and $F'(x) =f(x)$ for each $x in (a,b)$. Show that $f$ is measurable on $(a,b)$.
Hint : Show first that $forall x in (a,b) : f(x) = limlimits_{n to infty} n[F(x+frac{1}{n}) - F(x)]$.
I have already proved the hint ... How to proceed please?
measurable-functions
measurable-functions
asked Nov 23 at 22:01
Ahmed
1,244511
asked Nov 23 at 22:01
Ahmed
1,244511
asked Nov 23 at 22:01
Ahmed
1,244511
asked Nov 23 at 22:01
Ahmed
1,244511
asked Nov 23 at 22:01
Ahmed
1,244511
1,244511
add a comment |
add a comment |
1 Answer
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Note that each function $xmapsto n(F(x + 1/n) - F(x))$ is measurable and that the limit of a sequence of measurable functions is measurable.
answered Nov 23 at 22:04
ncmathsadist
41.9k259101
We can say this because : Since F is differentiable then it is continuous and so it is measurable. Isn't it?
– Ahmed
Nov 23 at 22:14
1
That is correct.
– ncmathsadist
Nov 23 at 22:17
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Note that each function $xmapsto n(F(x + 1/n) - F(x))$ is measurable and that the limit of a sequence of measurable functions is measurable.
answered Nov 23 at 22:04
ncmathsadist
41.9k259101
We can say this because : Since F is differentiable then it is continuous and so it is measurable. Isn't it?
– Ahmed
Nov 23 at 22:14
1
That is correct.
– ncmathsadist
Nov 23 at 22:17
add a comment |
up vote
0
down vote
accepted
Note that each function $xmapsto n(F(x + 1/n) - F(x))$ is measurable and that the limit of a sequence of measurable functions is measurable.
answered Nov 23 at 22:04
ncmathsadist
41.9k259101
We can say this because : Since F is differentiable then it is continuous and so it is measurable. Isn't it?
– Ahmed
Nov 23 at 22:14
1
That is correct.
– ncmathsadist
Nov 23 at 22:17
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Note that each function $xmapsto n(F(x + 1/n) - F(x))$ is measurable and that the limit of a sequence of measurable functions is measurable.
answered Nov 23 at 22:04
ncmathsadist
41.9k259101
Note that each function $xmapsto n(F(x + 1/n) - F(x))$ is measurable and that the limit of a sequence of measurable functions is measurable.
answered Nov 23 at 22:04
ncmathsadist
41.9k259101
answered Nov 23 at 22:04
ncmathsadist
41.9k259101
answered Nov 23 at 22:04
ncmathsadist
41.9k259101
answered Nov 23 at 22:04
ncmathsadist
41.9k259101
41.9k259101
We can say this because : Since F is differentiable then it is continuous and so it is measurable. Isn't it?
– Ahmed
Nov 23 at 22:14
1
That is correct.
– ncmathsadist
Nov 23 at 22:17
add a comment |
We can say this because : Since F is differentiable then it is continuous and so it is measurable. Isn't it?
– Ahmed
Nov 23 at 22:14
1
That is correct.
– ncmathsadist
Nov 23 at 22:17
We can say this because : Since F is differentiable then it is continuous and so it is measurable. Isn't it?
– Ahmed
Nov 23 at 22:14
We can say this because : Since F is differentiable then it is continuous and so it is measurable. Isn't it?
– Ahmed
Nov 23 at 22:14
1
1
That is correct.
– ncmathsadist
Nov 23 at 22:17
That is correct.
– ncmathsadist
Nov 23 at 22:17
add a comment |
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