On a compact support of a measure

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Let $mu$ be a measure on the Borel-$sigma$-field of $mathbbR$ such that $mu (mathbbR)=1$. Recall that the support of $mu$ is the largest closed set $C$ such that for all open sets $U$ with $Ucap C neq phi$, we have $mu(U)>0$. Assume that every continuous real-valued function on $mathbbR$ is integrable with respect to $mu$. Prove that the support of $mu$ is compact.

I think all we need to prove is that such a $C$ is bounded. Then by Heine-Borel we can conclude $C$ is compact. But merely the integrability of a continuous function w.r.t. $mu$ is not going to help me to show that $C$ is bounded. So, what will be the best possible approach for that problem?

probability-theory measure-theory compactness

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edited Jul 30 at 8:46

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Let $mu$ be a measure on the Borel-$sigma$-field of $mathbbR$ such that $mu (mathbbR)=1$. Recall that the support of $mu$ is the largest closed set $C$ such that for all open sets $U$ with $Ucap C neq phi$, we have $mu(U)>0$. Assume that every continuous real-valued function on $mathbbR$ is integrable with respect to $mu$. Prove that the support of $mu$ is compact.

I think all we need to prove is that such a $C$ is bounded. Then by Heine-Borel we can conclude $C$ is compact. But merely the integrability of a continuous function w.r.t. $mu$ is not going to help me to show that $C$ is bounded. So, what will be the best possible approach for that problem?

probability-theory measure-theory compactness

share|cite|improve this question

edited Jul 30 at 8:46

Davide Giraudo

121k15146249

asked Jul 27 at 16:52

am_11235...

1797

add a comment | 

up vote
3
down vote

favorite

up vote
3
down vote

favorite

Let $mu$ be a measure on the Borel-$sigma$-field of $mathbbR$ such that $mu (mathbbR)=1$. Recall that the support of $mu$ is the largest closed set $C$ such that for all open sets $U$ with $Ucap C neq phi$, we have $mu(U)>0$. Assume that every continuous real-valued function on $mathbbR$ is integrable with respect to $mu$. Prove that the support of $mu$ is compact.

I think all we need to prove is that such a $C$ is bounded. Then by Heine-Borel we can conclude $C$ is compact. But merely the integrability of a continuous function w.r.t. $mu$ is not going to help me to show that $C$ is bounded. So, what will be the best possible approach for that problem?

probability-theory measure-theory compactness

share|cite|improve this question

edited Jul 30 at 8:46

Davide Giraudo

121k15146249

asked Jul 27 at 16:52

am_11235...

1797

Let $mu$ be a measure on the Borel-$sigma$-field of $mathbbR$ such that $mu (mathbbR)=1$. Recall that the support of $mu$ is the largest closed set $C$ such that for all open sets $U$ with $Ucap C neq phi$, we have $mu(U)>0$. Assume that every continuous real-valued function on $mathbbR$ is integrable with respect to $mu$. Prove that the support of $mu$ is compact.

I think all we need to prove is that such a $C$ is bounded. Then by Heine-Borel we can conclude $C$ is compact. But merely the integrability of a continuous function w.r.t. $mu$ is not going to help me to show that $C$ is bounded. So, what will be the best possible approach for that problem?

probability-theory measure-theory compactness

share|cite|improve this question

edited Jul 30 at 8:46

Davide Giraudo

121k15146249

asked Jul 27 at 16:52

am_11235...

1797

share|cite|improve this question

share|cite|improve this question

edited Jul 30 at 8:46

Davide Giraudo

121k15146249

edited Jul 30 at 8:46

Davide Giraudo

121k15146249

edited Jul 30 at 8:46

Davide Giraudo

121k15146249

121k15146249

asked Jul 27 at 16:52

am_11235...

1797

asked Jul 27 at 16:52

am_11235...

1797

asked Jul 27 at 16:52

am_11235...

1797

1797

add a comment | 

add a comment | 

2 Answers
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If $operatornamesupp(mu)$ wasn’t bounded, we would be able to find a sequence $(x_n)$ with $lim_n x_n =infty$ and a sequence of disjoint neighborhoods $(N_n)$ of $x_n$ such that for all $n in mathbb N$, $mu(N_n)>0$.

We can find for each $n$ an interval $I_n$ of length $delta_n$ centered on $x_n$ and included in $N_n$. From there, for each $n$, we can build a continuous function with support included in $I_n$ and of integral greater than $1$. The sum $f$ of those functions converges as all their supports are disjoint. And the integral of $f$ is unbounded in contradiction with our hypothesis.

share|cite|improve this answer

edited Jul 30 at 8:46

Davide Giraudo

121k15146249

answered Jul 27 at 17:08

mathcounterexamples.net

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Let $left(a_nright)_ninmathbb Z$ be a sequence of positive real numbers. Let $f_n$ be a continuous function supported in $(n-1,n+2)$ such that $0leqslant f_n(x)leqslant a_n$ for all $x$ and $f_n(x)=a_n$ for $xin (n,n+1]$. Define $fcolon xmapsto sum_ninmathbb Zf_n(x)$ (the definition makes sense, since for all $x$, the numbers of terms in $sum_ninmathbb Zf_n(x)$ which do not vanish is at most $3$). Since the convergence is uniform on compact sets, the function $f$ is continuous hence
$$
+inftygt int_mathbb Rf(x)mathrm dmu(x)=sum_ninmathbb Zint_mathbb Rf_n(x)mathrm dmu(x)geqslant sum_ninmathbb Zint_(n,n+1]f_n(x)mathrm dmu(x).$$
This shows that for any sequence $left(a_nright)_ninmathbb Z$ of positive real numbers, the series $sum_ninmathbb Za_nmuleft((n,n+1]right)$ converges. A good choice allows to conclude.

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answered Jul 27 at 17:09

Davide Giraudo

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote

If $operatornamesupp(mu)$ wasn’t bounded, we would be able to find a sequence $(x_n)$ with $lim_n x_n =infty$ and a sequence of disjoint neighborhoods $(N_n)$ of $x_n$ such that for all $n in mathbb N$, $mu(N_n)>0$.

We can find for each $n$ an interval $I_n$ of length $delta_n$ centered on $x_n$ and included in $N_n$. From there, for each $n$, we can build a continuous function with support included in $I_n$ and of integral greater than $1$. The sum $f$ of those functions converges as all their supports are disjoint. And the integral of $f$ is unbounded in contradiction with our hypothesis.

share|cite|improve this answer

edited Jul 30 at 8:46

Davide Giraudo

121k15146249

answered Jul 27 at 17:08

mathcounterexamples.net

23.5k21652

add a comment | 

up vote
1
down vote

If $operatornamesupp(mu)$ wasn’t bounded, we would be able to find a sequence $(x_n)$ with $lim_n x_n =infty$ and a sequence of disjoint neighborhoods $(N_n)$ of $x_n$ such that for all $n in mathbb N$, $mu(N_n)>0$.

We can find for each $n$ an interval $I_n$ of length $delta_n$ centered on $x_n$ and included in $N_n$. From there, for each $n$, we can build a continuous function with support included in $I_n$ and of integral greater than $1$. The sum $f$ of those functions converges as all their supports are disjoint. And the integral of $f$ is unbounded in contradiction with our hypothesis.

share|cite|improve this answer

edited Jul 30 at 8:46

Davide Giraudo

121k15146249

answered Jul 27 at 17:08

mathcounterexamples.net

23.5k21652

add a comment | 

up vote
1
down vote

up vote
1
down vote

If $operatornamesupp(mu)$ wasn’t bounded, we would be able to find a sequence $(x_n)$ with $lim_n x_n =infty$ and a sequence of disjoint neighborhoods $(N_n)$ of $x_n$ such that for all $n in mathbb N$, $mu(N_n)>0$.

We can find for each $n$ an interval $I_n$ of length $delta_n$ centered on $x_n$ and included in $N_n$. From there, for each $n$, we can build a continuous function with support included in $I_n$ and of integral greater than $1$. The sum $f$ of those functions converges as all their supports are disjoint. And the integral of $f$ is unbounded in contradiction with our hypothesis.

share|cite|improve this answer

edited Jul 30 at 8:46

Davide Giraudo

121k15146249

answered Jul 27 at 17:08

mathcounterexamples.net

23.5k21652

If $operatornamesupp(mu)$ wasn’t bounded, we would be able to find a sequence $(x_n)$ with $lim_n x_n =infty$ and a sequence of disjoint neighborhoods $(N_n)$ of $x_n$ such that for all $n in mathbb N$, $mu(N_n)>0$.

We can find for each $n$ an interval $I_n$ of length $delta_n$ centered on $x_n$ and included in $N_n$. From there, for each $n$, we can build a continuous function with support included in $I_n$ and of integral greater than $1$. The sum $f$ of those functions converges as all their supports are disjoint. And the integral of $f$ is unbounded in contradiction with our hypothesis.

share|cite|improve this answer

edited Jul 30 at 8:46

Davide Giraudo

121k15146249

answered Jul 27 at 17:08

mathcounterexamples.net

23.5k21652

share|cite|improve this answer

share|cite|improve this answer

edited Jul 30 at 8:46

Davide Giraudo

121k15146249

edited Jul 30 at 8:46

Davide Giraudo

121k15146249

edited Jul 30 at 8:46

Davide Giraudo

121k15146249

121k15146249

answered Jul 27 at 17:08

mathcounterexamples.net

23.5k21652

answered Jul 27 at 17:08

mathcounterexamples.net

23.5k21652

answered Jul 27 at 17:08

mathcounterexamples.net

23.5k21652

23.5k21652

add a comment | 

add a comment | 

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0
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Let $left(a_nright)_ninmathbb Z$ be a sequence of positive real numbers. Let $f_n$ be a continuous function supported in $(n-1,n+2)$ such that $0leqslant f_n(x)leqslant a_n$ for all $x$ and $f_n(x)=a_n$ for $xin (n,n+1]$. Define $fcolon xmapsto sum_ninmathbb Zf_n(x)$ (the definition makes sense, since for all $x$, the numbers of terms in $sum_ninmathbb Zf_n(x)$ which do not vanish is at most $3$). Since the convergence is uniform on compact sets, the function $f$ is continuous hence
$$
+inftygt int_mathbb Rf(x)mathrm dmu(x)=sum_ninmathbb Zint_mathbb Rf_n(x)mathrm dmu(x)geqslant sum_ninmathbb Zint_(n,n+1]f_n(x)mathrm dmu(x).$$
This shows that for any sequence $left(a_nright)_ninmathbb Z$ of positive real numbers, the series $sum_ninmathbb Za_nmuleft((n,n+1]right)$ converges. A good choice allows to conclude.

share|cite|improve this answer

answered Jul 27 at 17:09

Davide Giraudo

121k15146249

add a comment | 

up vote
0
down vote

Let $left(a_nright)_ninmathbb Z$ be a sequence of positive real numbers. Let $f_n$ be a continuous function supported in $(n-1,n+2)$ such that $0leqslant f_n(x)leqslant a_n$ for all $x$ and $f_n(x)=a_n$ for $xin (n,n+1]$. Define $fcolon xmapsto sum_ninmathbb Zf_n(x)$ (the definition makes sense, since for all $x$, the numbers of terms in $sum_ninmathbb Zf_n(x)$ which do not vanish is at most $3$). Since the convergence is uniform on compact sets, the function $f$ is continuous hence
$$
+inftygt int_mathbb Rf(x)mathrm dmu(x)=sum_ninmathbb Zint_mathbb Rf_n(x)mathrm dmu(x)geqslant sum_ninmathbb Zint_(n,n+1]f_n(x)mathrm dmu(x).$$
This shows that for any sequence $left(a_nright)_ninmathbb Z$ of positive real numbers, the series $sum_ninmathbb Za_nmuleft((n,n+1]right)$ converges. A good choice allows to conclude.

share|cite|improve this answer

answered Jul 27 at 17:09

Davide Giraudo

121k15146249

add a comment | 

up vote
0
down vote

up vote
0
down vote

Let $left(a_nright)_ninmathbb Z$ be a sequence of positive real numbers. Let $f_n$ be a continuous function supported in $(n-1,n+2)$ such that $0leqslant f_n(x)leqslant a_n$ for all $x$ and $f_n(x)=a_n$ for $xin (n,n+1]$. Define $fcolon xmapsto sum_ninmathbb Zf_n(x)$ (the definition makes sense, since for all $x$, the numbers of terms in $sum_ninmathbb Zf_n(x)$ which do not vanish is at most $3$). Since the convergence is uniform on compact sets, the function $f$ is continuous hence
$$
+inftygt int_mathbb Rf(x)mathrm dmu(x)=sum_ninmathbb Zint_mathbb Rf_n(x)mathrm dmu(x)geqslant sum_ninmathbb Zint_(n,n+1]f_n(x)mathrm dmu(x).$$
This shows that for any sequence $left(a_nright)_ninmathbb Z$ of positive real numbers, the series $sum_ninmathbb Za_nmuleft((n,n+1]right)$ converges. A good choice allows to conclude.

share|cite|improve this answer

answered Jul 27 at 17:09

Davide Giraudo

121k15146249

Let $left(a_nright)_ninmathbb Z$ be a sequence of positive real numbers. Let $f_n$ be a continuous function supported in $(n-1,n+2)$ such that $0leqslant f_n(x)leqslant a_n$ for all $x$ and $f_n(x)=a_n$ for $xin (n,n+1]$. Define $fcolon xmapsto sum_ninmathbb Zf_n(x)$ (the definition makes sense, since for all $x$, the numbers of terms in $sum_ninmathbb Zf_n(x)$ which do not vanish is at most $3$). Since the convergence is uniform on compact sets, the function $f$ is continuous hence
$$
+inftygt int_mathbb Rf(x)mathrm dmu(x)=sum_ninmathbb Zint_mathbb Rf_n(x)mathrm dmu(x)geqslant sum_ninmathbb Zint_(n,n+1]f_n(x)mathrm dmu(x).$$
This shows that for any sequence $left(a_nright)_ninmathbb Z$ of positive real numbers, the series $sum_ninmathbb Za_nmuleft((n,n+1]right)$ converges. A good choice allows to conclude.

share|cite|improve this answer

answered Jul 27 at 17:09

Davide Giraudo

121k15146249

share|cite|improve this answer

share|cite|improve this answer

answered Jul 27 at 17:09

Davide Giraudo

121k15146249

answered Jul 27 at 17:09

Davide Giraudo

121k15146249

answered Jul 27 at 17:09

Davide Giraudo

121k15146249

121k15146249

add a comment | 

add a comment | 

 

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