If a probability measure has two densities, then they agree a.e. (Proof verification)
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Suppose that $f, g$ are two nonnegative measurable functions that integrate to one over $mathbf{R}$, with respect to Lebesgue measure and are such that the probability measure $mu(A) = int_A f dlambda = int_A g dlambda$ for all Borels $A$. Our goal is to show that $f = g$ a.e.
I think the proof can go like this. Let $E = {x : f(x) > g(x)}$, and suppose that $lambda(E) > 0$. Then for some $n$, $E_n = {x : f(x) - g(x) geq 1/n}$ is such that $lambda(E_n) > 0$ (by countable subadditivity this must be true), in which case $int_{E_n} (f - g) dlambda > 0$, which is a contradiction. We use here that $f$ and $g$ are measurable so that $E_n$ is Borel. Reversing the roles of $f$ and $g$, this means ${x : f(x) > g(x)} cup {x : g(x) > f(x)} = {x : f(x)neq g(x)}$ is null as desired.
By the way, it looks like the more general statement is that if $f$ and $g$ are two measurable functions agreeing (by integration) over all Borels, then $f = g$ a.e.
My only question is if this argument looks right.
probability-theory measure-theory proof-verification
asked Dec 8 '18 at 21:35
Drew BradyDrew Brady
719315
$endgroup$
add a comment |
$begingroup$
Suppose that $f, g$ are two nonnegative measurable functions that integrate to one over $mathbf{R}$, with respect to Lebesgue measure and are such that the probability measure $mu(A) = int_A f dlambda = int_A g dlambda$ for all Borels $A$. Our goal is to show that $f = g$ a.e.
I think the proof can go like this. Let $E = {x : f(x) > g(x)}$, and suppose that $lambda(E) > 0$. Then for some $n$, $E_n = {x : f(x) - g(x) geq 1/n}$ is such that $lambda(E_n) > 0$ (by countable subadditivity this must be true), in which case $int_{E_n} (f - g) dlambda > 0$, which is a contradiction. We use here that $f$ and $g$ are measurable so that $E_n$ is Borel. Reversing the roles of $f$ and $g$, this means ${x : f(x) > g(x)} cup {x : g(x) > f(x)} = {x : f(x)neq g(x)}$ is null as desired.
By the way, it looks like the more general statement is that if $f$ and $g$ are two measurable functions agreeing (by integration) over all Borels, then $f = g$ a.e.
My only question is if this argument looks right.
probability-theory measure-theory proof-verification
asked Dec 8 '18 at 21:35
Drew BradyDrew Brady
719315
$endgroup$
$begingroup$
Yes, the last statement is a well known fact that you just proved. By the way you don't need to check two cases $f(x)>g(x)$ and $g(x)>f(x)$ if you put an absolute value.
$endgroup$
– Yanko
Dec 8 '18 at 21:38
$begingroup$
Thanks @Yanko!!
$endgroup$
– Drew Brady
Dec 8 '18 at 21:45
$begingroup$
@Yanko The absolute value does not pass to the integral hence one very much needs to check the two cases (or to justify that the second case works like the first one, by symmetry).
$endgroup$
– Did
Dec 8 '18 at 21:47
$begingroup$
@Did You're right indeed. Idk why I said that.
$endgroup$
– Yanko
Dec 8 '18 at 21:49
$begingroup$
Yes, I tried to get it to work using the absolute value, but it seems not to work (but it may be possible in the special case that $f$ is integrating to one and nonnegative as in the pdf), but in general I don't think it can be made to work directly, as $int f < infty$ does not in general imply $int |f| < +infty$.
$endgroup$
– Drew Brady
Dec 8 '18 at 21:50
add a comment |
$begingroup$
Suppose that $f, g$ are two nonnegative measurable functions that integrate to one over $mathbf{R}$, with respect to Lebesgue measure and are such that the probability measure $mu(A) = int_A f dlambda = int_A g dlambda$ for all Borels $A$. Our goal is to show that $f = g$ a.e.
I think the proof can go like this. Let $E = {x : f(x) > g(x)}$, and suppose that $lambda(E) > 0$. Then for some $n$, $E_n = {x : f(x) - g(x) geq 1/n}$ is such that $lambda(E_n) > 0$ (by countable subadditivity this must be true), in which case $int_{E_n} (f - g) dlambda > 0$, which is a contradiction. We use here that $f$ and $g$ are measurable so that $E_n$ is Borel. Reversing the roles of $f$ and $g$, this means ${x : f(x) > g(x)} cup {x : g(x) > f(x)} = {x : f(x)neq g(x)}$ is null as desired.
By the way, it looks like the more general statement is that if $f$ and $g$ are two measurable functions agreeing (by integration) over all Borels, then $f = g$ a.e.
My only question is if this argument looks right.
probability-theory measure-theory proof-verification
asked Dec 8 '18 at 21:35
Drew BradyDrew Brady
719315
$endgroup$
Suppose that $f, g$ are two nonnegative measurable functions that integrate to one over $mathbf{R}$, with respect to Lebesgue measure and are such that the probability measure $mu(A) = int_A f dlambda = int_A g dlambda$ for all Borels $A$. Our goal is to show that $f = g$ a.e.
I think the proof can go like this. Let $E = {x : f(x) > g(x)}$, and suppose that $lambda(E) > 0$. Then for some $n$, $E_n = {x : f(x) - g(x) geq 1/n}$ is such that $lambda(E_n) > 0$ (by countable subadditivity this must be true), in which case $int_{E_n} (f - g) dlambda > 0$, which is a contradiction. We use here that $f$ and $g$ are measurable so that $E_n$ is Borel. Reversing the roles of $f$ and $g$, this means ${x : f(x) > g(x)} cup {x : g(x) > f(x)} = {x : f(x)neq g(x)}$ is null as desired.
By the way, it looks like the more general statement is that if $f$ and $g$ are two measurable functions agreeing (by integration) over all Borels, then $f = g$ a.e.
My only question is if this argument looks right.
probability-theory measure-theory proof-verification
probability-theory measure-theory proof-verification
asked Dec 8 '18 at 21:35
Drew BradyDrew Brady
719315
asked Dec 8 '18 at 21:35
Drew BradyDrew Brady
719315
edited Dec 8 '18 at 21:45
Drew Brady
asked Dec 8 '18 at 21:35
Drew BradyDrew Brady
719315
asked Dec 8 '18 at 21:35
Drew BradyDrew Brady
719315
asked Dec 8 '18 at 21:35
Drew BradyDrew Brady
719315
719315
$begingroup$
Yes, the last statement is a well known fact that you just proved. By the way you don't need to check two cases $f(x)>g(x)$ and $g(x)>f(x)$ if you put an absolute value.
$endgroup$
– Yanko
Dec 8 '18 at 21:38
$begingroup$
Thanks @Yanko!!
$endgroup$
– Drew Brady
Dec 8 '18 at 21:45
$begingroup$
@Yanko The absolute value does not pass to the integral hence one very much needs to check the two cases (or to justify that the second case works like the first one, by symmetry).
$endgroup$
– Did
Dec 8 '18 at 21:47
$begingroup$
@Did You're right indeed. Idk why I said that.
$endgroup$
– Yanko
Dec 8 '18 at 21:49
$begingroup$
Yes, I tried to get it to work using the absolute value, but it seems not to work (but it may be possible in the special case that $f$ is integrating to one and nonnegative as in the pdf), but in general I don't think it can be made to work directly, as $int f < infty$ does not in general imply $int |f| < +infty$.
$endgroup$
– Drew Brady
Dec 8 '18 at 21:50
add a comment |
$begingroup$
Yes, the last statement is a well known fact that you just proved. By the way you don't need to check two cases $f(x)>g(x)$ and $g(x)>f(x)$ if you put an absolute value.
$endgroup$
– Yanko
Dec 8 '18 at 21:38
$begingroup$
Thanks @Yanko!!
$endgroup$
– Drew Brady
Dec 8 '18 at 21:45
$begingroup$
@Yanko The absolute value does not pass to the integral hence one very much needs to check the two cases (or to justify that the second case works like the first one, by symmetry).
$endgroup$
– Did
Dec 8 '18 at 21:47
$begingroup$
@Did You're right indeed. Idk why I said that.
$endgroup$
– Yanko
Dec 8 '18 at 21:49
$begingroup$
Yes, I tried to get it to work using the absolute value, but it seems not to work (but it may be possible in the special case that $f$ is integrating to one and nonnegative as in the pdf), but in general I don't think it can be made to work directly, as $int f < infty$ does not in general imply $int |f| < +infty$.
$endgroup$
– Drew Brady
Dec 8 '18 at 21:50
$begingroup$
Yes, the last statement is a well known fact that you just proved. By the way you don't need to check two cases $f(x)>g(x)$ and $g(x)>f(x)$ if you put an absolute value.
$endgroup$
– Yanko
Dec 8 '18 at 21:38
$begingroup$
Yes, the last statement is a well known fact that you just proved. By the way you don't need to check two cases $f(x)>g(x)$ and $g(x)>f(x)$ if you put an absolute value.
$endgroup$
– Yanko
Dec 8 '18 at 21:38
$begingroup$
Thanks @Yanko!!
$endgroup$
– Drew Brady
Dec 8 '18 at 21:45
$begingroup$
Thanks @Yanko!!
$endgroup$
– Drew Brady
Dec 8 '18 at 21:45
$begingroup$
@Yanko The absolute value does not pass to the integral hence one very much needs to check the two cases (or to justify that the second case works like the first one, by symmetry).
$endgroup$
– Did
Dec 8 '18 at 21:47
$begingroup$
@Yanko The absolute value does not pass to the integral hence one very much needs to check the two cases (or to justify that the second case works like the first one, by symmetry).
$endgroup$
– Did
Dec 8 '18 at 21:47
$begingroup$
@Did You're right indeed. Idk why I said that.
$endgroup$
– Yanko
Dec 8 '18 at 21:49
$begingroup$
@Did You're right indeed. Idk why I said that.
$endgroup$
– Yanko
Dec 8 '18 at 21:49
$begingroup$
Yes, I tried to get it to work using the absolute value, but it seems not to work (but it may be possible in the special case that $f$ is integrating to one and nonnegative as in the pdf), but in general I don't think it can be made to work directly, as $int f < infty$ does not in general imply $int |f| < +infty$.
$endgroup$
– Drew Brady
Dec 8 '18 at 21:50
$begingroup$
Yes, I tried to get it to work using the absolute value, but it seems not to work (but it may be possible in the special case that $f$ is integrating to one and nonnegative as in the pdf), but in general I don't think it can be made to work directly, as $int f < infty$ does not in general imply $int |f| < +infty$.
$endgroup$
– Drew Brady
Dec 8 '18 at 21:50
add a comment |
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$begingroup$
Yes, the last statement is a well known fact that you just proved. By the way you don't need to check two cases $f(x)>g(x)$ and $g(x)>f(x)$ if you put an absolute value.
$endgroup$
– Yanko
Dec 8 '18 at 21:38
$begingroup$
Thanks @Yanko!!
$endgroup$
– Drew Brady
Dec 8 '18 at 21:45
$begingroup$
@Yanko The absolute value does not pass to the integral hence one very much needs to check the two cases (or to justify that the second case works like the first one, by symmetry).
$endgroup$
– Did
Dec 8 '18 at 21:47
$begingroup$
@Did You're right indeed. Idk why I said that.
$endgroup$
– Yanko
Dec 8 '18 at 21:49
$begingroup$
Yes, I tried to get it to work using the absolute value, but it seems not to work (but it may be possible in the special case that $f$ is integrating to one and nonnegative as in the pdf), but in general I don't think it can be made to work directly, as $int f < infty$ does not in general imply $int |f| < +infty$.
$endgroup$
– Drew Brady
Dec 8 '18 at 21:50