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$f: [0,1] to mathbbR$ absolutely continuous, $f' in 0,1$ (a.e.), $f(0)=0$. Prove that for some measurable subset $A$, $f(x)=m(A cap (0,x))$

Minimal dense subset of $mathbbQ cap [0,1]$Prove that $(B, |-|_infty)$ complete. B the set of bounded real valued functions on [0,1] which are pointwise limit of continuous functions.Result about Lebesgue measurable subset of $mathbbR^2$ and horizontal/vertical slicesProve/disprove if $f$ is continuous on $[0,1]$, and absolutely continuous on $(a,1], ain (0,1)$, $f$ is absolutely continuous on $[0,1]$.Let $f:ErightarrowmathbbR$ with $EinmathscrM$. $f$ is measurable iff for every open subset of $mathbbR$, $f^-1(A)$ is measurableProve that a.e. convergence with bounded integral implies $L^1$ convergeceSequence of measurable functions $g_n_n=1^infty$ such that $lim_ntoinftyint_0^1f(t)g_n(t)dt=int_0^1f(t)dt$ whenever $fin C[0,1]$Show that $limlimits_ntoinftyint_Asin(nx)dx = 0$ for any Lebesgue measurable $A subseteq [0,1]$Sequence $f_n$ of non-neg. meas. on [0,1] s.t $f_n to 0$ a.e but if $[a,b]subset[0,1]$ then $lim_ntoinftyint_a^bf_n(x)dx =(b-a)$If $f$ is absolutely continuous and $g$ is continuous, prove $f' =g$.

3

$begingroup$

Problem: Suppose that $f: [0,1] to mathbbR$ is absolutely continuous, $f' in 0,1$ (a.e.) and $f(0)=0$. Prove that for some measurable subset $A subset [0,1]$ and every $x in [0,1]$ we have $f(x)=m(A cap (0,x))$

I have proved it on my own. Please help me verify and tell me how much of the problem you think I have solved. (This was an exam problem and I'm trying to verify my given answer). I didn't have time to study derivatives for the real analysis exam and my argument might lack a bit of rigor, I think.

Edit: I'm going to write down my proof as an answer to mark this question solved.

real-analysis analysis measure-theory lebesgue-integral absolute-continuity

share|cite|improve this question

edited 15 hours ago

stressed out

asked Jun 23 '18 at 14:54

stressed outstressed out

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$endgroup$

add a comment | 

3

$begingroup$

Problem: Suppose that $f: [0,1] to mathbbR$ is absolutely continuous, $f' in 0,1$ (a.e.) and $f(0)=0$. Prove that for some measurable subset $A subset [0,1]$ and every $x in [0,1]$ we have $f(x)=m(A cap (0,x))$

I have proved it on my own. Please help me verify and tell me how much of the problem you think I have solved. (This was an exam problem and I'm trying to verify my given answer). I didn't have time to study derivatives for the real analysis exam and my argument might lack a bit of rigor, I think.

Edit: I'm going to write down my proof as an answer to mark this question solved.

real-analysis analysis measure-theory lebesgue-integral absolute-continuity

share|cite|improve this question

edited 15 hours ago

stressed out

asked Jun 23 '18 at 14:54

stressed outstressed out

6,5431939

$endgroup$

add a comment | 

$begingroup$

Problem: Suppose that $f: [0,1] to mathbbR$ is absolutely continuous, $f' in 0,1$ (a.e.) and $f(0)=0$. Prove that for some measurable subset $A subset [0,1]$ and every $x in [0,1]$ we have $f(x)=m(A cap (0,x))$

I have proved it on my own. Please help me verify and tell me how much of the problem you think I have solved. (This was an exam problem and I'm trying to verify my given answer). I didn't have time to study derivatives for the real analysis exam and my argument might lack a bit of rigor, I think.

Edit: I'm going to write down my proof as an answer to mark this question solved.

real-analysis analysis measure-theory lebesgue-integral absolute-continuity

share|cite|improve this question

edited 15 hours ago

stressed out

asked Jun 23 '18 at 14:54

stressed outstressed out

6,5431939

$endgroup$

share|cite|improve this question

edited 15 hours ago

stressed out

asked Jun 23 '18 at 14:54

stressed outstressed out

6,5431939

share|cite|improve this question

edited 15 hours ago

stressed out

asked Jun 23 '18 at 14:54

stressed outstressed out

6,5431939

asked Jun 23 '18 at 14:54

stressed outstressed out

6,5431939

asked Jun 23 '18 at 14:54

stressed outstressed out

6,5431939

1 Answer
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1

$begingroup$

Suppose that $f: [0,1] to mathbbR$ is absolutely continuous, $f' in 0,1$ (a.e.) and $f(0)=0$. Prove that for some measurable subset $A subset [0,1]$ and every $x in [0,1]$ we have $f(x)=m(A cap (0,x))$

Since $f$ is absolutely continuous, we know that it comes from an indefinite integral like below:

$$f(x) = f(0) + int_0^xf'$$

Set $A = xin [0,1] : f'(x) = 1$, $B = xin [0,1] : f'(x) = 0$ and $C = (Acup B)^c$. By assumption, $m(C)=0$. Moreover, since $f'$ is a measurable function (being the limit of the measurable function $mathrmDiff_1/n(f)$ when $n$ goes to infinity), $A$ and $B$ are measurable sets and $C$ is measurable because its measure is $0$ and we're working with Lebesgue measure.

Using the assumptions, we have:

$$f(x) = 0 + int_0^x f' times mathbf1_A + int_0^x f' times mathbf1_B + int_0^x f' times mathbf1_C$$

But the two integrals on the right will vanish and we'll get

$$f(x) = int_0^x1times mathbf1_A = m(A cap(0,x))$$

Remark: $mathbf1_X$ denotes the indicator function on the set $X$.

share|cite|improve this answer

answered 15 hours ago

stressed outstressed out

6,5431939

$endgroup$

add a comment | 

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1

$begingroup$

Suppose that $f: [0,1] to mathbbR$ is absolutely continuous, $f' in 0,1$ (a.e.) and $f(0)=0$. Prove that for some measurable subset $A subset [0,1]$ and every $x in [0,1]$ we have $f(x)=m(A cap (0,x))$

Since $f$ is absolutely continuous, we know that it comes from an indefinite integral like below:

$$f(x) = f(0) + int_0^xf'$$

Set $A = xin [0,1] : f'(x) = 1$, $B = xin [0,1] : f'(x) = 0$ and $C = (Acup B)^c$. By assumption, $m(C)=0$. Moreover, since $f'$ is a measurable function (being the limit of the measurable function $mathrmDiff_1/n(f)$ when $n$ goes to infinity), $A$ and $B$ are measurable sets and $C$ is measurable because its measure is $0$ and we're working with Lebesgue measure.

Using the assumptions, we have:

$$f(x) = 0 + int_0^x f' times mathbf1_A + int_0^x f' times mathbf1_B + int_0^x f' times mathbf1_C$$

But the two integrals on the right will vanish and we'll get

$$f(x) = int_0^x1times mathbf1_A = m(A cap(0,x))$$

Remark: $mathbf1_X$ denotes the indicator function on the set $X$.

share|cite|improve this answer

answered 15 hours ago

stressed outstressed out

6,5431939

$endgroup$

add a comment | 

1

$begingroup$

Suppose that $f: [0,1] to mathbbR$ is absolutely continuous, $f' in 0,1$ (a.e.) and $f(0)=0$. Prove that for some measurable subset $A subset [0,1]$ and every $x in [0,1]$ we have $f(x)=m(A cap (0,x))$

Since $f$ is absolutely continuous, we know that it comes from an indefinite integral like below:

$$f(x) = f(0) + int_0^xf'$$

Set $A = xin [0,1] : f'(x) = 1$, $B = xin [0,1] : f'(x) = 0$ and $C = (Acup B)^c$. By assumption, $m(C)=0$. Moreover, since $f'$ is a measurable function (being the limit of the measurable function $mathrmDiff_1/n(f)$ when $n$ goes to infinity), $A$ and $B$ are measurable sets and $C$ is measurable because its measure is $0$ and we're working with Lebesgue measure.

Using the assumptions, we have:

$$f(x) = 0 + int_0^x f' times mathbf1_A + int_0^x f' times mathbf1_B + int_0^x f' times mathbf1_C$$

But the two integrals on the right will vanish and we'll get

$$f(x) = int_0^x1times mathbf1_A = m(A cap(0,x))$$

Remark: $mathbf1_X$ denotes the indicator function on the set $X$.

share|cite|improve this answer

answered 15 hours ago

stressed outstressed out

6,5431939

$endgroup$

add a comment | 

$begingroup$

Suppose that $f: [0,1] to mathbbR$ is absolutely continuous, $f' in 0,1$ (a.e.) and $f(0)=0$. Prove that for some measurable subset $A subset [0,1]$ and every $x in [0,1]$ we have $f(x)=m(A cap (0,x))$

Since $f$ is absolutely continuous, we know that it comes from an indefinite integral like below:

$$f(x) = f(0) + int_0^xf'$$

Set $A = xin [0,1] : f'(x) = 1$, $B = xin [0,1] : f'(x) = 0$ and $C = (Acup B)^c$. By assumption, $m(C)=0$. Moreover, since $f'$ is a measurable function (being the limit of the measurable function $mathrmDiff_1/n(f)$ when $n$ goes to infinity), $A$ and $B$ are measurable sets and $C$ is measurable because its measure is $0$ and we're working with Lebesgue measure.

Using the assumptions, we have:

$$f(x) = 0 + int_0^x f' times mathbf1_A + int_0^x f' times mathbf1_B + int_0^x f' times mathbf1_C$$

But the two integrals on the right will vanish and we'll get

$$f(x) = int_0^x1times mathbf1_A = m(A cap(0,x))$$

Remark: $mathbf1_X$ denotes the indicator function on the set $X$.

share|cite|improve this answer

answered 15 hours ago

stressed outstressed out

6,5431939

$endgroup$

share|cite|improve this answer

answered 15 hours ago

stressed outstressed out

6,5431939

answered 15 hours ago

stressed outstressed out

6,5431939

answered 15 hours ago

stressed outstressed out

6,5431939