I have a measure &#x03BC;<!-- μ --> that is supported on [ &#x2212;<!-- − --> 3 ,

Poftethef9t

Poftethef9t

Answered question

2022-06-16

I have a measure μ that is supported on [ 3 , 3 ] × R . What we are given is that, if we fix the first component i, then μ ( i , ) is a probability measure. Formally (maybe it should integrate to d i or similar, I cannot define this very well):
x R d μ ( i , x ) = 1 i [ 3 , 3 ] .
I find it very hard to understand this tuple-indexing notation. My question is the following: What kind of assumptions do we need to have a result similar to:
( i , x ) [ 3 , 3 ] × R d μ ( i , x ) = i [ 3 , 3 ] d i

I just think that since for any fixed i the measure μ integrates to 1 (on the second dimension -- apologies for my poor terminology), then integrating over all ( i , x ) [ 3 , 3 ] × R should also give simply an 'iterated integral' where we first integrate wrt x R for a fixed i, which will integrate to 1 , and then integrate over i [ 3 , 3 ]. But of course, we cannot define
( i , x ) [ 3 , 3 ] × R d μ ( i , x ) = i [ 3 , 3 ] x R d μ ( i , x ) = i [ 3 , 3 ] d y
where in the red parts I make an abuse of integartion rules.

Answer & Explanation

arhaitategr

arhaitategr

Beginner2022-06-17Added 13 answers

The first equation you wrote, I think you probably mean μ ( { i } × R ) = 1?
What you are trying to describe with iterated integrals might be the product measure and the Fubini/Tonelli theorems which allow you to convert an integral over a product measure ν 1 × ν 2 to an iterated integral.
Finley Mckinney

Finley Mckinney

Beginner2022-06-18Added 11 answers

[ 3 , 3 ] × R d μ ( i , x ) = μ ( [ 3 , 3 ] × R ) by definition.

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