MathJax(?): Can't find handler for document MathJax(?): Can't find handler for document Let a &#x2208; X and &#x03BC; be a measure defined on 2 X by

∑ y i μ ( A i ) = y i if there i is such that a ∈ A i and 0 if there is no such i. Note that if A i 's are disjoint there can be at most one i for which A ∈ A i Hnece ∫ s d μ = s ( a ). From this it follows that ∫ f d μ = f ( a ) for all non-negative measurable functiosn f. Hence, the conditioin is f ( a ) &lt; ∞.

Expert Answer to "MathJax(?): Can't find handler for document MathJax(?): Can't..."

Wade Bullock

Answered question

2022-07-10

MathJax(?): Can't find handler for document MathJax(?): Can't find handler for document Let

a \in X

and

μ

be a measure defined on

2^{X}

μ (E) = {\begin{cases} 1, & a \in E \\ 0, & a \notin E \end{cases}

Write a simple necessary and sufficient condition on the non-negative functions

f

that ensures that

\int_{X} f d μ < \infty

My attempt:
We know that

\int_{X} f d μ = sup {\int_{X} s d μ = \sum y_{i} μ (A_{i}) ∣ s simple, s \geq 0, s \leq f}

, so we want the supremum to be finite and for this I'm thinking that the functions should have a finite border but I'm not sure if it works!

Answer & Explanation

Tanner Hamilton

Beginner2022-07-11Added 12 answers

\sum y_{i} μ (A_{i}) = y_{i}

if there

i

is such that

a \in A_{i}

and 0 if there is no such

i

. Note that if

A_{i}

's are disjoint there can be at most one i for which

A \in A_{i}

Hnece

\int s d μ = s (a)

. From this it follows that

\int f d μ = f (a)

for all non-negative measurable functiosn

f

. Hence, the conditioin is

f (a) < \infty

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