Mixing
If a function is integrable, does it also have finite integral given any counting measure?
$begingroup$
If the integral of a function $f$ is $L^1$,
begin{equation} int f dx < infty end{equation}
Does the same function have finite integral under any countable counting measure,
begin{equation} sum f(x_{i}) dmu(x_{i}) < infty end{equation}
for any sequance ${ x_{i} }_{i in mathbb{N}}$?
I wanna say yes, due to the latter being the same as simple functions on point sets. But also we might loose cancelation of areas when throwing away alot of stuff in this fashion.
real-analysis
asked Jan 1 at 16:12
user7534user7534
645
$endgroup$
add a comment |
$begingroup$
If the integral of a function $f$ is $L^1$,
begin{equation} int f dx < infty end{equation}
Does the same function have finite integral under any countable counting measure,
begin{equation} sum f(x_{i}) dmu(x_{i}) < infty end{equation}
for any sequance ${ x_{i} }_{i in mathbb{N}}$?
I wanna say yes, due to the latter being the same as simple functions on point sets. But also we might loose cancelation of areas when throwing away alot of stuff in this fashion.
real-analysis
asked Jan 1 at 16:12
user7534user7534
645
$endgroup$
4
$begingroup$
I don't think that $sum f(x_i)dmu (x_i) $ is a standard notation
$endgroup$
– Jakobian
Jan 1 at 16:16
$begingroup$
@Jakobian it is any counting measure with weight 1 on each point.
$endgroup$
– user7534
Jan 1 at 16:20
3
$begingroup$
@user7534 Then you should write $int fdmu$ or $sum_{i} f(x_i)mu({x_i})$. Not a mixup of both.
$endgroup$
– drhab
Jan 1 at 16:29
add a comment |
$begingroup$
If the integral of a function $f$ is $L^1$,
begin{equation} int f dx < infty end{equation}
Does the same function have finite integral under any countable counting measure,
begin{equation} sum f(x_{i}) dmu(x_{i}) < infty end{equation}
for any sequance ${ x_{i} }_{i in mathbb{N}}$?
I wanna say yes, due to the latter being the same as simple functions on point sets. But also we might loose cancelation of areas when throwing away alot of stuff in this fashion.
real-analysis
asked Jan 1 at 16:12
user7534user7534
645
$endgroup$
If the integral of a function $f$ is $L^1$,
begin{equation} int f dx < infty end{equation}
Does the same function have finite integral under any countable counting measure,
begin{equation} sum f(x_{i}) dmu(x_{i}) < infty end{equation}
for any sequance ${ x_{i} }_{i in mathbb{N}}$?
I wanna say yes, due to the latter being the same as simple functions on point sets. But also we might loose cancelation of areas when throwing away alot of stuff in this fashion.
real-analysis
real-analysis
asked Jan 1 at 16:12
user7534user7534
645
asked Jan 1 at 16:12
user7534user7534
645
edited Jan 1 at 16:26
user7534
asked Jan 1 at 16:12
user7534user7534
645
asked Jan 1 at 16:12
user7534user7534
645
asked Jan 1 at 16:12
user7534user7534
645
645
4
$begingroup$
I don't think that $sum f(x_i)dmu (x_i) $ is a standard notation
$endgroup$
– Jakobian
Jan 1 at 16:16
$begingroup$
@Jakobian it is any counting measure with weight 1 on each point.
$endgroup$
– user7534
Jan 1 at 16:20
3
$begingroup$
@user7534 Then you should write $int fdmu$ or $sum_{i} f(x_i)mu({x_i})$. Not a mixup of both.
$endgroup$
– drhab
Jan 1 at 16:29
add a comment |
4
$begingroup$
I don't think that $sum f(x_i)dmu (x_i) $ is a standard notation
$endgroup$
– Jakobian
Jan 1 at 16:16
$begingroup$
@Jakobian it is any counting measure with weight 1 on each point.
$endgroup$
– user7534
Jan 1 at 16:20
3
$begingroup$
@user7534 Then you should write $int fdmu$ or $sum_{i} f(x_i)mu({x_i})$. Not a mixup of both.
$endgroup$
– drhab
Jan 1 at 16:29
4
4
$begingroup$
I don't think that $sum f(x_i)dmu (x_i) $ is a standard notation
$endgroup$
– Jakobian
Jan 1 at 16:16
$begingroup$
I don't think that $sum f(x_i)dmu (x_i) $ is a standard notation
$endgroup$
– Jakobian
Jan 1 at 16:16
$begingroup$
@Jakobian it is any counting measure with weight 1 on each point.
$endgroup$
– user7534
Jan 1 at 16:20
$begingroup$
@Jakobian it is any counting measure with weight 1 on each point.
$endgroup$
– user7534
Jan 1 at 16:20
3
3
$begingroup$
@user7534 Then you should write $int fdmu$ or $sum_{i} f(x_i)mu({x_i})$. Not a mixup of both.
$endgroup$
– drhab
Jan 1 at 16:29
$begingroup$
@user7534 Then you should write $int fdmu$ or $sum_{i} f(x_i)mu({x_i})$. Not a mixup of both.
$endgroup$
– drhab
Jan 1 at 16:29
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
So, first of all, please make sure you get your notation correct, otherwise it's hard to be sure what you're asking. I think you mean to say that $fin L_1$, not $L_2$. It would also be nice to have the domain of $f$, which I assume from context is $(0,infty)$. Also, if you want to talk about an integral over the counting measure on $mathbb{N}$, the proper notation is
$$sum_{i=1}^infty f(x_i).$$
Of course, the answer is no. Consider the function $f=boldsymbol{1}_mathbb{N}$.
There are special conditions under which the answer is yes. By the Integral Test, if $f$ is nonnegative, continuous, and decreasing on $(0,infty)$ and $int f<infty$ then $sum_{n=1}^infty f(n)$ converges. In fact, so does $sum_{i=1}^infty f(x_i)$, provided $(x_i)_{i=1}^infty$ is increasing with $inf|x_{i+1}-x_i|>0$.
answered Jan 1 at 16:24
Ben WBen W
2,115615
$endgroup$
$begingroup$
Thanks, you mind clarify the last part with a lower bound on the increment?
$endgroup$
– user7534
Jan 1 at 16:33
1
$begingroup$
@user7534 Sure. The gritty details are similar to the proof of the Integral Test itself (see wikipedia for an outline), but the underlying idea is quite intuitive. Suppose each $x_iin(i,i+1]$. Then $f(x_i)leq f(i)$ and so $sum f(x_i)leq sum f(i)$, the latter of which we already decided converges. Of course, it doesn't matter if we scale the $x_i$'s or cut out some of them.
$endgroup$
– Ben W
Jan 1 at 16:38
add a comment |
$begingroup$
If $f:mathbb Rtomathbb R$ is defined as $f=mathbf1_{mathbb Z}$ then $int f(x)^2dx=0$ but $sum_{ninmathbb Z}f(n)=infty$.
answered Jan 1 at 16:21
drhabdrhab
98.6k544129
$endgroup$
add a comment |
$begingroup$
Your question is confusing, but I think the answer is essentially, "no".
You can change the value of an integrable function on any countable set of points without changing integrability or the integral. So just define it to be large enough at each point at which your "counting measure" is supported to make the sum diverge.
answered Jan 1 at 16:19
Ethan BolkerEthan Bolker
42.1k548111
$endgroup$
add a comment |
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3 Answers
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oldest
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3 Answers
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$begingroup$
So, first of all, please make sure you get your notation correct, otherwise it's hard to be sure what you're asking. I think you mean to say that $fin L_1$, not $L_2$. It would also be nice to have the domain of $f$, which I assume from context is $(0,infty)$. Also, if you want to talk about an integral over the counting measure on $mathbb{N}$, the proper notation is
$$sum_{i=1}^infty f(x_i).$$
Of course, the answer is no. Consider the function $f=boldsymbol{1}_mathbb{N}$.
There are special conditions under which the answer is yes. By the Integral Test, if $f$ is nonnegative, continuous, and decreasing on $(0,infty)$ and $int f<infty$ then $sum_{n=1}^infty f(n)$ converges. In fact, so does $sum_{i=1}^infty f(x_i)$, provided $(x_i)_{i=1}^infty$ is increasing with $inf|x_{i+1}-x_i|>0$.
answered Jan 1 at 16:24
Ben WBen W
2,115615
$endgroup$
$begingroup$
Thanks, you mind clarify the last part with a lower bound on the increment?
$endgroup$
– user7534
Jan 1 at 16:33
1
$begingroup$
@user7534 Sure. The gritty details are similar to the proof of the Integral Test itself (see wikipedia for an outline), but the underlying idea is quite intuitive. Suppose each $x_iin(i,i+1]$. Then $f(x_i)leq f(i)$ and so $sum f(x_i)leq sum f(i)$, the latter of which we already decided converges. Of course, it doesn't matter if we scale the $x_i$'s or cut out some of them.
$endgroup$
– Ben W
Jan 1 at 16:38
add a comment |
$begingroup$
So, first of all, please make sure you get your notation correct, otherwise it's hard to be sure what you're asking. I think you mean to say that $fin L_1$, not $L_2$. It would also be nice to have the domain of $f$, which I assume from context is $(0,infty)$. Also, if you want to talk about an integral over the counting measure on $mathbb{N}$, the proper notation is
$$sum_{i=1}^infty f(x_i).$$
Of course, the answer is no. Consider the function $f=boldsymbol{1}_mathbb{N}$.
There are special conditions under which the answer is yes. By the Integral Test, if $f$ is nonnegative, continuous, and decreasing on $(0,infty)$ and $int f<infty$ then $sum_{n=1}^infty f(n)$ converges. In fact, so does $sum_{i=1}^infty f(x_i)$, provided $(x_i)_{i=1}^infty$ is increasing with $inf|x_{i+1}-x_i|>0$.
answered Jan 1 at 16:24
Ben WBen W
2,115615
$endgroup$
$begingroup$
Thanks, you mind clarify the last part with a lower bound on the increment?
$endgroup$
– user7534
Jan 1 at 16:33
1
$begingroup$
@user7534 Sure. The gritty details are similar to the proof of the Integral Test itself (see wikipedia for an outline), but the underlying idea is quite intuitive. Suppose each $x_iin(i,i+1]$. Then $f(x_i)leq f(i)$ and so $sum f(x_i)leq sum f(i)$, the latter of which we already decided converges. Of course, it doesn't matter if we scale the $x_i$'s or cut out some of them.
$endgroup$
– Ben W
Jan 1 at 16:38
add a comment |
$begingroup$
So, first of all, please make sure you get your notation correct, otherwise it's hard to be sure what you're asking. I think you mean to say that $fin L_1$, not $L_2$. It would also be nice to have the domain of $f$, which I assume from context is $(0,infty)$. Also, if you want to talk about an integral over the counting measure on $mathbb{N}$, the proper notation is
$$sum_{i=1}^infty f(x_i).$$
Of course, the answer is no. Consider the function $f=boldsymbol{1}_mathbb{N}$.
There are special conditions under which the answer is yes. By the Integral Test, if $f$ is nonnegative, continuous, and decreasing on $(0,infty)$ and $int f<infty$ then $sum_{n=1}^infty f(n)$ converges. In fact, so does $sum_{i=1}^infty f(x_i)$, provided $(x_i)_{i=1}^infty$ is increasing with $inf|x_{i+1}-x_i|>0$.
answered Jan 1 at 16:24
Ben WBen W
2,115615
$endgroup$
So, first of all, please make sure you get your notation correct, otherwise it's hard to be sure what you're asking. I think you mean to say that $fin L_1$, not $L_2$. It would also be nice to have the domain of $f$, which I assume from context is $(0,infty)$. Also, if you want to talk about an integral over the counting measure on $mathbb{N}$, the proper notation is
$$sum_{i=1}^infty f(x_i).$$
Of course, the answer is no. Consider the function $f=boldsymbol{1}_mathbb{N}$.
There are special conditions under which the answer is yes. By the Integral Test, if $f$ is nonnegative, continuous, and decreasing on $(0,infty)$ and $int f<infty$ then $sum_{n=1}^infty f(n)$ converges. In fact, so does $sum_{i=1}^infty f(x_i)$, provided $(x_i)_{i=1}^infty$ is increasing with $inf|x_{i+1}-x_i|>0$.
answered Jan 1 at 16:24
Ben WBen W
2,115615
edited Jan 1 at 16:43
answered Jan 1 at 16:24
Ben WBen W
2,115615
answered Jan 1 at 16:24
Ben WBen W
2,115615
answered Jan 1 at 16:24
Ben WBen W
2,115615
2,115615
$begingroup$
Thanks, you mind clarify the last part with a lower bound on the increment?
$endgroup$
– user7534
Jan 1 at 16:33
1
$begingroup$
@user7534 Sure. The gritty details are similar to the proof of the Integral Test itself (see wikipedia for an outline), but the underlying idea is quite intuitive. Suppose each $x_iin(i,i+1]$. Then $f(x_i)leq f(i)$ and so $sum f(x_i)leq sum f(i)$, the latter of which we already decided converges. Of course, it doesn't matter if we scale the $x_i$'s or cut out some of them.
$endgroup$
– Ben W
Jan 1 at 16:38
add a comment |
$begingroup$
Thanks, you mind clarify the last part with a lower bound on the increment?
$endgroup$
– user7534
Jan 1 at 16:33
1
$begingroup$
@user7534 Sure. The gritty details are similar to the proof of the Integral Test itself (see wikipedia for an outline), but the underlying idea is quite intuitive. Suppose each $x_iin(i,i+1]$. Then $f(x_i)leq f(i)$ and so $sum f(x_i)leq sum f(i)$, the latter of which we already decided converges. Of course, it doesn't matter if we scale the $x_i$'s or cut out some of them.
$endgroup$
– Ben W
Jan 1 at 16:38
$begingroup$
Thanks, you mind clarify the last part with a lower bound on the increment?
$endgroup$
– user7534
Jan 1 at 16:33
$begingroup$
Thanks, you mind clarify the last part with a lower bound on the increment?
$endgroup$
– user7534
Jan 1 at 16:33
1
1
$begingroup$
@user7534 Sure. The gritty details are similar to the proof of the Integral Test itself (see wikipedia for an outline), but the underlying idea is quite intuitive. Suppose each $x_iin(i,i+1]$. Then $f(x_i)leq f(i)$ and so $sum f(x_i)leq sum f(i)$, the latter of which we already decided converges. Of course, it doesn't matter if we scale the $x_i$'s or cut out some of them.
$endgroup$
– Ben W
Jan 1 at 16:38
$begingroup$
@user7534 Sure. The gritty details are similar to the proof of the Integral Test itself (see wikipedia for an outline), but the underlying idea is quite intuitive. Suppose each $x_iin(i,i+1]$. Then $f(x_i)leq f(i)$ and so $sum f(x_i)leq sum f(i)$, the latter of which we already decided converges. Of course, it doesn't matter if we scale the $x_i$'s or cut out some of them.
$endgroup$
– Ben W
Jan 1 at 16:38
add a comment |
$begingroup$
If $f:mathbb Rtomathbb R$ is defined as $f=mathbf1_{mathbb Z}$ then $int f(x)^2dx=0$ but $sum_{ninmathbb Z}f(n)=infty$.
answered Jan 1 at 16:21
drhabdrhab
98.6k544129
$endgroup$
add a comment |
$begingroup$
If $f:mathbb Rtomathbb R$ is defined as $f=mathbf1_{mathbb Z}$ then $int f(x)^2dx=0$ but $sum_{ninmathbb Z}f(n)=infty$.
answered Jan 1 at 16:21
drhabdrhab
98.6k544129
$endgroup$
add a comment |
$begingroup$
If $f:mathbb Rtomathbb R$ is defined as $f=mathbf1_{mathbb Z}$ then $int f(x)^2dx=0$ but $sum_{ninmathbb Z}f(n)=infty$.
answered Jan 1 at 16:21
drhabdrhab
98.6k544129
$endgroup$
If $f:mathbb Rtomathbb R$ is defined as $f=mathbf1_{mathbb Z}$ then $int f(x)^2dx=0$ but $sum_{ninmathbb Z}f(n)=infty$.
answered Jan 1 at 16:21
drhabdrhab
98.6k544129
answered Jan 1 at 16:21
drhabdrhab
98.6k544129
answered Jan 1 at 16:21
drhabdrhab
98.6k544129
answered Jan 1 at 16:21
drhabdrhab
98.6k544129
98.6k544129
add a comment |
add a comment |
$begingroup$
Your question is confusing, but I think the answer is essentially, "no".
You can change the value of an integrable function on any countable set of points without changing integrability or the integral. So just define it to be large enough at each point at which your "counting measure" is supported to make the sum diverge.
answered Jan 1 at 16:19
Ethan BolkerEthan Bolker
42.1k548111
$endgroup$
add a comment |
$begingroup$
Your question is confusing, but I think the answer is essentially, "no".
You can change the value of an integrable function on any countable set of points without changing integrability or the integral. So just define it to be large enough at each point at which your "counting measure" is supported to make the sum diverge.
answered Jan 1 at 16:19
Ethan BolkerEthan Bolker
42.1k548111
$endgroup$
add a comment |
$begingroup$
Your question is confusing, but I think the answer is essentially, "no".
You can change the value of an integrable function on any countable set of points without changing integrability or the integral. So just define it to be large enough at each point at which your "counting measure" is supported to make the sum diverge.
answered Jan 1 at 16:19
Ethan BolkerEthan Bolker
42.1k548111
$endgroup$
Your question is confusing, but I think the answer is essentially, "no".
You can change the value of an integrable function on any countable set of points without changing integrability or the integral. So just define it to be large enough at each point at which your "counting measure" is supported to make the sum diverge.
answered Jan 1 at 16:19
Ethan BolkerEthan Bolker
42.1k548111
answered Jan 1 at 16:19
Ethan BolkerEthan Bolker
42.1k548111
answered Jan 1 at 16:19
Ethan BolkerEthan Bolker
42.1k548111
answered Jan 1 at 16:19
Ethan BolkerEthan Bolker
42.1k548111
42.1k548111
add a comment |
add a comment |
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$begingroup$
I don't think that $sum f(x_i)dmu (x_i) $ is a standard notation
$endgroup$
– Jakobian
Jan 1 at 16:16
$begingroup$
@Jakobian it is any counting measure with weight 1 on each point.
$endgroup$
– user7534
Jan 1 at 16:20
3
$begingroup$
@user7534 Then you should write $int fdmu$ or $sum_{i} f(x_i)mu({x_i})$. Not a mixup of both.
$endgroup$
– drhab
Jan 1 at 16:29