dimension of vector space of measurable functions over finite set X
$begingroup$
Suppose we have $X={1,2,3}$ and $M$ a $sigma$-algebra over X.
I've stumped across the question of finding the dimension of the vector space of all the $M$-measurable real functions over $X$ (e.g.$quad f:X to mathbb{R}$).
For the best of me i can't seem to wrap my head around this, even if i think it's rather intuitive.
My difficulty is coming up with a possible basis of functions for the vector space of functions from $X$ to $mathbb{R}$. Once i get that checking for misurability is trivial in this case, but i don't know how to approach basis of functions.
Of course functions from $X$ to $mathbb{R}$ have at most 3 distinct values. I thought of using as basis 3 separate functions, one for each value of $X$: $e_i(i)=1quad i=1,2,3$.
But i don't think this cuts it since i can't come up with a way of writing every f via linear combination of these basis functions.
Any hint on how to tackle this problem? Or maybe a general idea of how a base function should look like?
Thanks!
functions vector-spaces measurable-functions
asked Jan 12 at 17:31
WhiteEyeTreeWhiteEyeTree
105
$endgroup$
add a comment |
$begingroup$
Suppose we have $X={1,2,3}$ and $M$ a $sigma$-algebra over X.
I've stumped across the question of finding the dimension of the vector space of all the $M$-measurable real functions over $X$ (e.g.$quad f:X to mathbb{R}$).
For the best of me i can't seem to wrap my head around this, even if i think it's rather intuitive.
My difficulty is coming up with a possible basis of functions for the vector space of functions from $X$ to $mathbb{R}$. Once i get that checking for misurability is trivial in this case, but i don't know how to approach basis of functions.
Of course functions from $X$ to $mathbb{R}$ have at most 3 distinct values. I thought of using as basis 3 separate functions, one for each value of $X$: $e_i(i)=1quad i=1,2,3$.
But i don't think this cuts it since i can't come up with a way of writing every f via linear combination of these basis functions.
Any hint on how to tackle this problem? Or maybe a general idea of how a base function should look like?
Thanks!
functions vector-spaces measurable-functions
asked Jan 12 at 17:31
WhiteEyeTreeWhiteEyeTree
105
$endgroup$
add a comment |
$begingroup$
Suppose we have $X={1,2,3}$ and $M$ a $sigma$-algebra over X.
I've stumped across the question of finding the dimension of the vector space of all the $M$-measurable real functions over $X$ (e.g.$quad f:X to mathbb{R}$).
For the best of me i can't seem to wrap my head around this, even if i think it's rather intuitive.
My difficulty is coming up with a possible basis of functions for the vector space of functions from $X$ to $mathbb{R}$. Once i get that checking for misurability is trivial in this case, but i don't know how to approach basis of functions.
Of course functions from $X$ to $mathbb{R}$ have at most 3 distinct values. I thought of using as basis 3 separate functions, one for each value of $X$: $e_i(i)=1quad i=1,2,3$.
But i don't think this cuts it since i can't come up with a way of writing every f via linear combination of these basis functions.
Any hint on how to tackle this problem? Or maybe a general idea of how a base function should look like?
Thanks!
functions vector-spaces measurable-functions
asked Jan 12 at 17:31
WhiteEyeTreeWhiteEyeTree
105
$endgroup$
Suppose we have $X={1,2,3}$ and $M$ a $sigma$-algebra over X.
I've stumped across the question of finding the dimension of the vector space of all the $M$-measurable real functions over $X$ (e.g.$quad f:X to mathbb{R}$).
For the best of me i can't seem to wrap my head around this, even if i think it's rather intuitive.
My difficulty is coming up with a possible basis of functions for the vector space of functions from $X$ to $mathbb{R}$. Once i get that checking for misurability is trivial in this case, but i don't know how to approach basis of functions.
Of course functions from $X$ to $mathbb{R}$ have at most 3 distinct values. I thought of using as basis 3 separate functions, one for each value of $X$: $e_i(i)=1quad i=1,2,3$.
But i don't think this cuts it since i can't come up with a way of writing every f via linear combination of these basis functions.
Any hint on how to tackle this problem? Or maybe a general idea of how a base function should look like?
Thanks!
functions vector-spaces measurable-functions
functions vector-spaces measurable-functions
asked Jan 12 at 17:31
WhiteEyeTreeWhiteEyeTree
105
asked Jan 12 at 17:31
WhiteEyeTreeWhiteEyeTree
105
asked Jan 12 at 17:31
WhiteEyeTreeWhiteEyeTree
105
asked Jan 12 at 17:31
WhiteEyeTreeWhiteEyeTree
105
asked Jan 12 at 17:31
WhiteEyeTreeWhiteEyeTree
105
105
add a comment |
add a comment |
1 Answer
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$begingroup$
The dimension depends on $M$. If $M={X,emptyset}$ then measurable functions are constant, so the dimension is $1$. If, as you seem to be assuming, $M$ is the power set of $X$ then every function is measurable. In that case if you define $$e_x(y)=begin{cases}1,&(y=x),\0,&(yne x).end{cases}$$then yes, the functions $e_x$ for $xin X$ form a basis.
Hint for that: $(2,3,7)=2(1,0,0)+3(0,1,0)+7(0,0,1)$.
answered Jan 12 at 17:40
David C. UllrichDavid C. Ullrich
61.8k44095
$endgroup$
$begingroup$
If we assume $M={X, emptyset, {1}, {2,3}}$ then this has dimension 2 with $(1,0,0), (0,1,1)$ as basis using your vectorial notation. Am i correct?
$endgroup$
– WhiteEyeTree
Jan 12 at 18:06
$begingroup$
@WhiteEyeTree Yes.
$endgroup$
– David C. Ullrich
Jan 12 at 18:07
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The dimension depends on $M$. If $M={X,emptyset}$ then measurable functions are constant, so the dimension is $1$. If, as you seem to be assuming, $M$ is the power set of $X$ then every function is measurable. In that case if you define $$e_x(y)=begin{cases}1,&(y=x),\0,&(yne x).end{cases}$$then yes, the functions $e_x$ for $xin X$ form a basis.
Hint for that: $(2,3,7)=2(1,0,0)+3(0,1,0)+7(0,0,1)$.
answered Jan 12 at 17:40
David C. UllrichDavid C. Ullrich
61.8k44095
$endgroup$
$begingroup$
If we assume $M={X, emptyset, {1}, {2,3}}$ then this has dimension 2 with $(1,0,0), (0,1,1)$ as basis using your vectorial notation. Am i correct?
$endgroup$
– WhiteEyeTree
Jan 12 at 18:06
$begingroup$
@WhiteEyeTree Yes.
$endgroup$
– David C. Ullrich
Jan 12 at 18:07
add a comment |
$begingroup$
The dimension depends on $M$. If $M={X,emptyset}$ then measurable functions are constant, so the dimension is $1$. If, as you seem to be assuming, $M$ is the power set of $X$ then every function is measurable. In that case if you define $$e_x(y)=begin{cases}1,&(y=x),\0,&(yne x).end{cases}$$then yes, the functions $e_x$ for $xin X$ form a basis.
Hint for that: $(2,3,7)=2(1,0,0)+3(0,1,0)+7(0,0,1)$.
answered Jan 12 at 17:40
David C. UllrichDavid C. Ullrich
61.8k44095
$endgroup$
$begingroup$
If we assume $M={X, emptyset, {1}, {2,3}}$ then this has dimension 2 with $(1,0,0), (0,1,1)$ as basis using your vectorial notation. Am i correct?
$endgroup$
– WhiteEyeTree
Jan 12 at 18:06
$begingroup$
@WhiteEyeTree Yes.
$endgroup$
– David C. Ullrich
Jan 12 at 18:07
add a comment |
$begingroup$
The dimension depends on $M$. If $M={X,emptyset}$ then measurable functions are constant, so the dimension is $1$. If, as you seem to be assuming, $M$ is the power set of $X$ then every function is measurable. In that case if you define $$e_x(y)=begin{cases}1,&(y=x),\0,&(yne x).end{cases}$$then yes, the functions $e_x$ for $xin X$ form a basis.
Hint for that: $(2,3,7)=2(1,0,0)+3(0,1,0)+7(0,0,1)$.
answered Jan 12 at 17:40
David C. UllrichDavid C. Ullrich
61.8k44095
$endgroup$
The dimension depends on $M$. If $M={X,emptyset}$ then measurable functions are constant, so the dimension is $1$. If, as you seem to be assuming, $M$ is the power set of $X$ then every function is measurable. In that case if you define $$e_x(y)=begin{cases}1,&(y=x),\0,&(yne x).end{cases}$$then yes, the functions $e_x$ for $xin X$ form a basis.
Hint for that: $(2,3,7)=2(1,0,0)+3(0,1,0)+7(0,0,1)$.
answered Jan 12 at 17:40
David C. UllrichDavid C. Ullrich
61.8k44095
answered Jan 12 at 17:40
David C. UllrichDavid C. Ullrich
61.8k44095
answered Jan 12 at 17:40
David C. UllrichDavid C. Ullrich
61.8k44095
answered Jan 12 at 17:40
David C. UllrichDavid C. Ullrich
61.8k44095
61.8k44095
$begingroup$
If we assume $M={X, emptyset, {1}, {2,3}}$ then this has dimension 2 with $(1,0,0), (0,1,1)$ as basis using your vectorial notation. Am i correct?
$endgroup$
– WhiteEyeTree
Jan 12 at 18:06
$begingroup$
@WhiteEyeTree Yes.
$endgroup$
– David C. Ullrich
Jan 12 at 18:07
add a comment |
$begingroup$
If we assume $M={X, emptyset, {1}, {2,3}}$ then this has dimension 2 with $(1,0,0), (0,1,1)$ as basis using your vectorial notation. Am i correct?
$endgroup$
– WhiteEyeTree
Jan 12 at 18:06
$begingroup$
@WhiteEyeTree Yes.
$endgroup$
– David C. Ullrich
Jan 12 at 18:07
$begingroup$
If we assume $M={X, emptyset, {1}, {2,3}}$ then this has dimension 2 with $(1,0,0), (0,1,1)$ as basis using your vectorial notation. Am i correct?
$endgroup$
– WhiteEyeTree
Jan 12 at 18:06
$begingroup$
If we assume $M={X, emptyset, {1}, {2,3}}$ then this has dimension 2 with $(1,0,0), (0,1,1)$ as basis using your vectorial notation. Am i correct?
$endgroup$
– WhiteEyeTree
Jan 12 at 18:06
$begingroup$
@WhiteEyeTree Yes.
$endgroup$
– David C. Ullrich
Jan 12 at 18:07
$begingroup$
@WhiteEyeTree Yes.
$endgroup$
– David C. Ullrich
Jan 12 at 18:07
add a comment |
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