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Continuity under a measure integral
Existence of measure factorizationLeibniz's Derivative Rule for Integral in Measure TheoryContinuity under the integral sign?Absolute continuity under the integralLebesgue measure of region under curveDerivative of a probability measureStandard notation or name for the measure defined by an integral over a functionOn the continuity of a function given by the 'partial' Lebesgue integral of a multivariable functionImage measure- equivalent integration and integrableAbsolute continuity of probability measure wrt Lebesgue measure
$begingroup$
Suppose $(Omega, mathcal{F}, nu)$ is a measure space and that $f=f(x,t) : Omega times mathbb{R} to mathbb{R}$ is integrable and measurable in $x$ and is continuous in $t$. Define $F: mathbb{R} to mathbb{R}$ by
$$F(t) = int_{Omega} f(x,t) dnu(x).$$
Is $F$ continuous?
real-analysis measure-theory
asked Mar 15 at 9:34
talfredtalfred
1007
$endgroup$
add a comment |
$begingroup$
Suppose $(Omega, mathcal{F}, nu)$ is a measure space and that $f=f(x,t) : Omega times mathbb{R} to mathbb{R}$ is integrable and measurable in $x$ and is continuous in $t$. Define $F: mathbb{R} to mathbb{R}$ by
$$F(t) = int_{Omega} f(x,t) dnu(x).$$
Is $F$ continuous?
real-analysis measure-theory
asked Mar 15 at 9:34
talfredtalfred
1007
$endgroup$
1
$begingroup$
"Fundamental Theorem of Calculus' has a different meaning.
$endgroup$
– Kavi Rama Murthy
Mar 15 at 9:48
add a comment |
$begingroup$
Suppose $(Omega, mathcal{F}, nu)$ is a measure space and that $f=f(x,t) : Omega times mathbb{R} to mathbb{R}$ is integrable and measurable in $x$ and is continuous in $t$. Define $F: mathbb{R} to mathbb{R}$ by
$$F(t) = int_{Omega} f(x,t) dnu(x).$$
Is $F$ continuous?
real-analysis measure-theory
asked Mar 15 at 9:34
talfredtalfred
1007
$endgroup$
Suppose $(Omega, mathcal{F}, nu)$ is a measure space and that $f=f(x,t) : Omega times mathbb{R} to mathbb{R}$ is integrable and measurable in $x$ and is continuous in $t$. Define $F: mathbb{R} to mathbb{R}$ by
$$F(t) = int_{Omega} f(x,t) dnu(x).$$
Is $F$ continuous?
real-analysis measure-theory
real-analysis measure-theory
asked Mar 15 at 9:34
talfredtalfred
1007
asked Mar 15 at 9:34
talfredtalfred
1007
edited Mar 15 at 9:51
talfred
asked Mar 15 at 9:34
talfredtalfred
1007
asked Mar 15 at 9:34
talfredtalfred
1007
asked Mar 15 at 9:34
talfredtalfred
1007
1007
1
$begingroup$
"Fundamental Theorem of Calculus' has a different meaning.
$endgroup$
– Kavi Rama Murthy
Mar 15 at 9:48
add a comment |
1
$begingroup$
"Fundamental Theorem of Calculus' has a different meaning.
$endgroup$
– Kavi Rama Murthy
Mar 15 at 9:48
1
1
$begingroup$
"Fundamental Theorem of Calculus' has a different meaning.
$endgroup$
– Kavi Rama Murthy
Mar 15 at 9:48
$begingroup$
"Fundamental Theorem of Calculus' has a different meaning.
$endgroup$
– Kavi Rama Murthy
Mar 15 at 9:48
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $f(x,t)=sum g_n(t)I_{E_n}(x)$ where $E_n$'s are pairwise disjoint measurable sets and each $g_n$ is continuous. Then the hypothesis is satisfied and $F(t)=sum g_n(t)nu (E_n)$. So to get a counterexample, all you have to do is find continuous functions $g_n$ and positive numbers $a_n$ such that $sum a_ng_n$ converges but the sum is not continuous. (We can always find disjoint measurable sets $E_n$ with $a_n=nu (E_n)$. We can take $nu$ to be Lebesgue measure on $mathbb R$, for example). I will leave this part to you.
answered Mar 15 at 9:45
Kavi Rama MurthyKavi Rama Murthy
70.2k53170
$endgroup$
$begingroup$
So if that sum converges uniformly, I think we are okay? If we further assume that $|f(x,t)| leq g(x)$ for every $x, t$ for some integrable function $g$, I think we can appeal to the Dominated Convergence Theorem to get continuity?
$endgroup$
– talfred
Mar 15 at 9:55
1
$begingroup$
That is a good point. We need something like DCT to prove continuity of $F$ so we have to put some extra conditions. Uniform convergence will do the job if $nu$ is a finite measure.
$endgroup$
– Kavi Rama Murthy
Mar 15 at 9:57
$begingroup$
If you can bound $f$ by an integrable function, the result is true by DCT.
$endgroup$
– Math_QED
Mar 15 at 11:10
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Let $f(x,t)=sum g_n(t)I_{E_n}(x)$ where $E_n$'s are pairwise disjoint measurable sets and each $g_n$ is continuous. Then the hypothesis is satisfied and $F(t)=sum g_n(t)nu (E_n)$. So to get a counterexample, all you have to do is find continuous functions $g_n$ and positive numbers $a_n$ such that $sum a_ng_n$ converges but the sum is not continuous. (We can always find disjoint measurable sets $E_n$ with $a_n=nu (E_n)$. We can take $nu$ to be Lebesgue measure on $mathbb R$, for example). I will leave this part to you.
answered Mar 15 at 9:45
Kavi Rama MurthyKavi Rama Murthy
70.2k53170
$endgroup$
$begingroup$
So if that sum converges uniformly, I think we are okay? If we further assume that $|f(x,t)| leq g(x)$ for every $x, t$ for some integrable function $g$, I think we can appeal to the Dominated Convergence Theorem to get continuity?
$endgroup$
– talfred
Mar 15 at 9:55
1
$begingroup$
That is a good point. We need something like DCT to prove continuity of $F$ so we have to put some extra conditions. Uniform convergence will do the job if $nu$ is a finite measure.
$endgroup$
– Kavi Rama Murthy
Mar 15 at 9:57
$begingroup$
If you can bound $f$ by an integrable function, the result is true by DCT.
$endgroup$
– Math_QED
Mar 15 at 11:10
add a comment |
$begingroup$
Let $f(x,t)=sum g_n(t)I_{E_n}(x)$ where $E_n$'s are pairwise disjoint measurable sets and each $g_n$ is continuous. Then the hypothesis is satisfied and $F(t)=sum g_n(t)nu (E_n)$. So to get a counterexample, all you have to do is find continuous functions $g_n$ and positive numbers $a_n$ such that $sum a_ng_n$ converges but the sum is not continuous. (We can always find disjoint measurable sets $E_n$ with $a_n=nu (E_n)$. We can take $nu$ to be Lebesgue measure on $mathbb R$, for example). I will leave this part to you.
answered Mar 15 at 9:45
Kavi Rama MurthyKavi Rama Murthy
70.2k53170
$endgroup$
$begingroup$
So if that sum converges uniformly, I think we are okay? If we further assume that $|f(x,t)| leq g(x)$ for every $x, t$ for some integrable function $g$, I think we can appeal to the Dominated Convergence Theorem to get continuity?
$endgroup$
– talfred
Mar 15 at 9:55
1
$begingroup$
That is a good point. We need something like DCT to prove continuity of $F$ so we have to put some extra conditions. Uniform convergence will do the job if $nu$ is a finite measure.
$endgroup$
– Kavi Rama Murthy
Mar 15 at 9:57
$begingroup$
If you can bound $f$ by an integrable function, the result is true by DCT.
$endgroup$
– Math_QED
Mar 15 at 11:10
add a comment |
$begingroup$
Let $f(x,t)=sum g_n(t)I_{E_n}(x)$ where $E_n$'s are pairwise disjoint measurable sets and each $g_n$ is continuous. Then the hypothesis is satisfied and $F(t)=sum g_n(t)nu (E_n)$. So to get a counterexample, all you have to do is find continuous functions $g_n$ and positive numbers $a_n$ such that $sum a_ng_n$ converges but the sum is not continuous. (We can always find disjoint measurable sets $E_n$ with $a_n=nu (E_n)$. We can take $nu$ to be Lebesgue measure on $mathbb R$, for example). I will leave this part to you.
answered Mar 15 at 9:45
Kavi Rama MurthyKavi Rama Murthy
70.2k53170
$endgroup$
Let $f(x,t)=sum g_n(t)I_{E_n}(x)$ where $E_n$'s are pairwise disjoint measurable sets and each $g_n$ is continuous. Then the hypothesis is satisfied and $F(t)=sum g_n(t)nu (E_n)$. So to get a counterexample, all you have to do is find continuous functions $g_n$ and positive numbers $a_n$ such that $sum a_ng_n$ converges but the sum is not continuous. (We can always find disjoint measurable sets $E_n$ with $a_n=nu (E_n)$. We can take $nu$ to be Lebesgue measure on $mathbb R$, for example). I will leave this part to you.
answered Mar 15 at 9:45
Kavi Rama MurthyKavi Rama Murthy
70.2k53170
answered Mar 15 at 9:45
Kavi Rama MurthyKavi Rama Murthy
70.2k53170
answered Mar 15 at 9:45
Kavi Rama MurthyKavi Rama Murthy
70.2k53170
answered Mar 15 at 9:45
Kavi Rama MurthyKavi Rama Murthy
70.2k53170
70.2k53170
$begingroup$
So if that sum converges uniformly, I think we are okay? If we further assume that $|f(x,t)| leq g(x)$ for every $x, t$ for some integrable function $g$, I think we can appeal to the Dominated Convergence Theorem to get continuity?
$endgroup$
– talfred
Mar 15 at 9:55
1
$begingroup$
That is a good point. We need something like DCT to prove continuity of $F$ so we have to put some extra conditions. Uniform convergence will do the job if $nu$ is a finite measure.
$endgroup$
– Kavi Rama Murthy
Mar 15 at 9:57
$begingroup$
If you can bound $f$ by an integrable function, the result is true by DCT.
$endgroup$
– Math_QED
Mar 15 at 11:10
add a comment |
$begingroup$
So if that sum converges uniformly, I think we are okay? If we further assume that $|f(x,t)| leq g(x)$ for every $x, t$ for some integrable function $g$, I think we can appeal to the Dominated Convergence Theorem to get continuity?
$endgroup$
– talfred
Mar 15 at 9:55
1
$begingroup$
That is a good point. We need something like DCT to prove continuity of $F$ so we have to put some extra conditions. Uniform convergence will do the job if $nu$ is a finite measure.
$endgroup$
– Kavi Rama Murthy
Mar 15 at 9:57
$begingroup$
If you can bound $f$ by an integrable function, the result is true by DCT.
$endgroup$
– Math_QED
Mar 15 at 11:10
$begingroup$
So if that sum converges uniformly, I think we are okay? If we further assume that $|f(x,t)| leq g(x)$ for every $x, t$ for some integrable function $g$, I think we can appeal to the Dominated Convergence Theorem to get continuity?
$endgroup$
– talfred
Mar 15 at 9:55
$begingroup$
So if that sum converges uniformly, I think we are okay? If we further assume that $|f(x,t)| leq g(x)$ for every $x, t$ for some integrable function $g$, I think we can appeal to the Dominated Convergence Theorem to get continuity?
$endgroup$
– talfred
Mar 15 at 9:55
1
1
$begingroup$
That is a good point. We need something like DCT to prove continuity of $F$ so we have to put some extra conditions. Uniform convergence will do the job if $nu$ is a finite measure.
$endgroup$
– Kavi Rama Murthy
Mar 15 at 9:57
$begingroup$
That is a good point. We need something like DCT to prove continuity of $F$ so we have to put some extra conditions. Uniform convergence will do the job if $nu$ is a finite measure.
$endgroup$
– Kavi Rama Murthy
Mar 15 at 9:57
$begingroup$
If you can bound $f$ by an integrable function, the result is true by DCT.
$endgroup$
– Math_QED
Mar 15 at 11:10
$begingroup$
If you can bound $f$ by an integrable function, the result is true by DCT.
$endgroup$
– Math_QED
Mar 15 at 11:10
add a comment |
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$begingroup$
"Fundamental Theorem of Calculus' has a different meaning.
$endgroup$
– Kavi Rama Murthy
Mar 15 at 9:48