Mixing
Calculus notation in Haar measure. What is really going on?
$begingroup$
Let $G$ be a locally compact group with left Haar measure $mu$. If $f: G rightarrow mathbb{C}$ is an integrable function, and $E subseteq G$, the notation
$$int_E f(g)dg$$
is commonly used instead of $$intlimits_E f dmu$$
It is also common to write things like $intlimits_E f(g)d(gh^{-1})$ for a fixed $h in H$. This seems to be somewhat like u-substitution, although there should be some formal measure theory behind it. Would anyone be willing to explain or give a reference which explains the measure-theoretic principles behind the use of such notation?
real-analysis measure-theory topological-groups
asked Nov 11 '16 at 22:50
D_SD_S
13.9k61553
$endgroup$
add a comment |
$begingroup$
Let $G$ be a locally compact group with left Haar measure $mu$. If $f: G rightarrow mathbb{C}$ is an integrable function, and $E subseteq G$, the notation
$$int_E f(g)dg$$
is commonly used instead of $$intlimits_E f dmu$$
It is also common to write things like $intlimits_E f(g)d(gh^{-1})$ for a fixed $h in H$. This seems to be somewhat like u-substitution, although there should be some formal measure theory behind it. Would anyone be willing to explain or give a reference which explains the measure-theoretic principles behind the use of such notation?
real-analysis measure-theory topological-groups
asked Nov 11 '16 at 22:50
D_SD_S
13.9k61553
$endgroup$
add a comment |
$begingroup$
Let $G$ be a locally compact group with left Haar measure $mu$. If $f: G rightarrow mathbb{C}$ is an integrable function, and $E subseteq G$, the notation
$$int_E f(g)dg$$
is commonly used instead of $$intlimits_E f dmu$$
It is also common to write things like $intlimits_E f(g)d(gh^{-1})$ for a fixed $h in H$. This seems to be somewhat like u-substitution, although there should be some formal measure theory behind it. Would anyone be willing to explain or give a reference which explains the measure-theoretic principles behind the use of such notation?
real-analysis measure-theory topological-groups
asked Nov 11 '16 at 22:50
D_SD_S
13.9k61553
$endgroup$
Let $G$ be a locally compact group with left Haar measure $mu$. If $f: G rightarrow mathbb{C}$ is an integrable function, and $E subseteq G$, the notation
$$int_E f(g)dg$$
is commonly used instead of $$intlimits_E f dmu$$
It is also common to write things like $intlimits_E f(g)d(gh^{-1})$ for a fixed $h in H$. This seems to be somewhat like u-substitution, although there should be some formal measure theory behind it. Would anyone be willing to explain or give a reference which explains the measure-theoretic principles behind the use of such notation?
real-analysis measure-theory topological-groups
real-analysis measure-theory topological-groups
asked Nov 11 '16 at 22:50
D_SD_S
13.9k61553
asked Nov 11 '16 at 22:50
D_SD_S
13.9k61553
asked Nov 11 '16 at 22:50
D_SD_S
13.9k61553
asked Nov 11 '16 at 22:50
D_SD_S
13.9k61553
asked Nov 11 '16 at 22:50
D_SD_S
13.9k61553
13.9k61553
add a comment |
add a comment |
1 Answer
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$begingroup$
Here is an elaboration on the first 2 pages of chapter 1 of Rudin's "Fourier Analysis on Groups". Points 1-3 being theory, 4 being an issue of convention.
On a locally compact group $G$, there is a Haar measure $m$ that is unique up to a positive multiplicative constant.
For example: if G is compact, impose the condition $m(G)=1$ for any Haar measure $m$. Since $m(A)=lambda m'(A)$ for any Borel set $A$, where $m$ and $m'$ are 2 Haar measures on $G$, this means we impose $lambda = 1$ and hence $m$ is unique.
3a. What if $G$ is locally compact (and not just compact)? Now pick a compact set $K$ and impose the condition $m(K)=1$. By the above case, this determines a unique Haar measure when restricted to K. By translational invariance this determines a unique Haar measure on the whole of $G$.
3b. In making the final sentence precise, there are a few subtleties. First, because of local compactness, we have a compact neighbourhood $K$ of the identity, hence $cup_x (K+x)$ covers $G$. Second, whenever $K + x_1$ and $K + x_2$ have non-empty intersection, the measures have to coincide, again by translational invariance.
3c. The canonical example (from wikipedia) is the topological group $(mathbb{R}, +)$ where we pick the interval $[0,1]$ to have measure 1.
- Now that we have established a unique measure $m$. The rest is notation. In measure theory notation, we have $int_G f dm = int_G f(x) dm(x)$. Having determined a unique measure, we then write $int_G f(x) dx$ to mean $int_G f(x) dm(x)$.
The final point is also answered here, in a slightly different context. Notation involving the Lebesgue integral.
answered Jan 24 at 23:44
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$begingroup$
Here is an elaboration on the first 2 pages of chapter 1 of Rudin's "Fourier Analysis on Groups". Points 1-3 being theory, 4 being an issue of convention.
On a locally compact group $G$, there is a Haar measure $m$ that is unique up to a positive multiplicative constant.
For example: if G is compact, impose the condition $m(G)=1$ for any Haar measure $m$. Since $m(A)=lambda m'(A)$ for any Borel set $A$, where $m$ and $m'$ are 2 Haar measures on $G$, this means we impose $lambda = 1$ and hence $m$ is unique.
3a. What if $G$ is locally compact (and not just compact)? Now pick a compact set $K$ and impose the condition $m(K)=1$. By the above case, this determines a unique Haar measure when restricted to K. By translational invariance this determines a unique Haar measure on the whole of $G$.
3b. In making the final sentence precise, there are a few subtleties. First, because of local compactness, we have a compact neighbourhood $K$ of the identity, hence $cup_x (K+x)$ covers $G$. Second, whenever $K + x_1$ and $K + x_2$ have non-empty intersection, the measures have to coincide, again by translational invariance.
3c. The canonical example (from wikipedia) is the topological group $(mathbb{R}, +)$ where we pick the interval $[0,1]$ to have measure 1.
- Now that we have established a unique measure $m$. The rest is notation. In measure theory notation, we have $int_G f dm = int_G f(x) dm(x)$. Having determined a unique measure, we then write $int_G f(x) dx$ to mean $int_G f(x) dm(x)$.
The final point is also answered here, in a slightly different context. Notation involving the Lebesgue integral.
answered Jan 24 at 23:44
strikeemblemstrikeemblem
215
$endgroup$
add a comment |
$begingroup$
Here is an elaboration on the first 2 pages of chapter 1 of Rudin's "Fourier Analysis on Groups". Points 1-3 being theory, 4 being an issue of convention.
On a locally compact group $G$, there is a Haar measure $m$ that is unique up to a positive multiplicative constant.
For example: if G is compact, impose the condition $m(G)=1$ for any Haar measure $m$. Since $m(A)=lambda m'(A)$ for any Borel set $A$, where $m$ and $m'$ are 2 Haar measures on $G$, this means we impose $lambda = 1$ and hence $m$ is unique.
3a. What if $G$ is locally compact (and not just compact)? Now pick a compact set $K$ and impose the condition $m(K)=1$. By the above case, this determines a unique Haar measure when restricted to K. By translational invariance this determines a unique Haar measure on the whole of $G$.
3b. In making the final sentence precise, there are a few subtleties. First, because of local compactness, we have a compact neighbourhood $K$ of the identity, hence $cup_x (K+x)$ covers $G$. Second, whenever $K + x_1$ and $K + x_2$ have non-empty intersection, the measures have to coincide, again by translational invariance.
3c. The canonical example (from wikipedia) is the topological group $(mathbb{R}, +)$ where we pick the interval $[0,1]$ to have measure 1.
- Now that we have established a unique measure $m$. The rest is notation. In measure theory notation, we have $int_G f dm = int_G f(x) dm(x)$. Having determined a unique measure, we then write $int_G f(x) dx$ to mean $int_G f(x) dm(x)$.
The final point is also answered here, in a slightly different context. Notation involving the Lebesgue integral.
answered Jan 24 at 23:44
strikeemblemstrikeemblem
215
$endgroup$
add a comment |
$begingroup$
Here is an elaboration on the first 2 pages of chapter 1 of Rudin's "Fourier Analysis on Groups". Points 1-3 being theory, 4 being an issue of convention.
On a locally compact group $G$, there is a Haar measure $m$ that is unique up to a positive multiplicative constant.
For example: if G is compact, impose the condition $m(G)=1$ for any Haar measure $m$. Since $m(A)=lambda m'(A)$ for any Borel set $A$, where $m$ and $m'$ are 2 Haar measures on $G$, this means we impose $lambda = 1$ and hence $m$ is unique.
3a. What if $G$ is locally compact (and not just compact)? Now pick a compact set $K$ and impose the condition $m(K)=1$. By the above case, this determines a unique Haar measure when restricted to K. By translational invariance this determines a unique Haar measure on the whole of $G$.
3b. In making the final sentence precise, there are a few subtleties. First, because of local compactness, we have a compact neighbourhood $K$ of the identity, hence $cup_x (K+x)$ covers $G$. Second, whenever $K + x_1$ and $K + x_2$ have non-empty intersection, the measures have to coincide, again by translational invariance.
3c. The canonical example (from wikipedia) is the topological group $(mathbb{R}, +)$ where we pick the interval $[0,1]$ to have measure 1.
- Now that we have established a unique measure $m$. The rest is notation. In measure theory notation, we have $int_G f dm = int_G f(x) dm(x)$. Having determined a unique measure, we then write $int_G f(x) dx$ to mean $int_G f(x) dm(x)$.
The final point is also answered here, in a slightly different context. Notation involving the Lebesgue integral.
answered Jan 24 at 23:44
strikeemblemstrikeemblem
215
$endgroup$
Here is an elaboration on the first 2 pages of chapter 1 of Rudin's "Fourier Analysis on Groups". Points 1-3 being theory, 4 being an issue of convention.
On a locally compact group $G$, there is a Haar measure $m$ that is unique up to a positive multiplicative constant.
For example: if G is compact, impose the condition $m(G)=1$ for any Haar measure $m$. Since $m(A)=lambda m'(A)$ for any Borel set $A$, where $m$ and $m'$ are 2 Haar measures on $G$, this means we impose $lambda = 1$ and hence $m$ is unique.
3a. What if $G$ is locally compact (and not just compact)? Now pick a compact set $K$ and impose the condition $m(K)=1$. By the above case, this determines a unique Haar measure when restricted to K. By translational invariance this determines a unique Haar measure on the whole of $G$.
3b. In making the final sentence precise, there are a few subtleties. First, because of local compactness, we have a compact neighbourhood $K$ of the identity, hence $cup_x (K+x)$ covers $G$. Second, whenever $K + x_1$ and $K + x_2$ have non-empty intersection, the measures have to coincide, again by translational invariance.
3c. The canonical example (from wikipedia) is the topological group $(mathbb{R}, +)$ where we pick the interval $[0,1]$ to have measure 1.
- Now that we have established a unique measure $m$. The rest is notation. In measure theory notation, we have $int_G f dm = int_G f(x) dm(x)$. Having determined a unique measure, we then write $int_G f(x) dx$ to mean $int_G f(x) dm(x)$.
The final point is also answered here, in a slightly different context. Notation involving the Lebesgue integral.
answered Jan 24 at 23:44
strikeemblemstrikeemblem
215
edited Jan 25 at 0:00
answered Jan 24 at 23:44
strikeemblemstrikeemblem
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answered Jan 24 at 23:44
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215
answered Jan 24 at 23:44
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