A measure related with Reimann integral

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Homework Statement


Let [itex]( \mathbb{R}^k , \mathcal{A} , m_{k} )[/itex] be a Lebesgue measurable space, i.e., [itex]m_{k}=m[/itex] is a Lebesgue measure. Let [itex]f: \mathbb{R^k} \to \mathbb{R}[/itex] be a [itex]m[/itex]-integrable function. Define a function [itex]\mu : \mathcal{A} \to [0,\infty][/itex] by $$ \mu(A) := \int_{A} f(x) dx $$ with [itex]A \in \mathcal{A}[/itex]. Then is [itex]\mu[/itex] a measure?

Qeustion. Is there any error in the below working? And does it need any additional hypothesis to show $mu$ is a measure? Or is it acutally not a measure?

Homework Equations





The Attempt at a Solution


When [itex]A[/itex] is an empty set, the integral is obviously 0. Now suppose [itex]A_{0} , \dots , A_{n} \in \mathcal{A}[/itex] as they are disjoint. Then $$ \mu ( \bigcup_{k=0}^{n} A_{k} ) := \int_{\mathbb{R}^{k}} \chi_{\bigcup_{k=0}^{n} A_{k}}(x)f(x)dx $$ and we get $$ \chi_{\bigcup_{k=0}^{n} A_{k}} = \sum_{k=0}^{n} \chi_{A_k}$$ as [itex]A_{k}[/itex] are disjoint. Therefore, we have $$ \mu ( \bigcup_{k=0}^{n} A_{k} ) = \sum_{k=0}^{n} \mu(A_{k}). $$ Thus taking [itex]n \to \infty[/itex] we get the required result.
 

Answers and Replies

  • #2
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Homework Statement


Let [itex]( \mathbb{R}^k , \mathcal{A} , m_{k} )[/itex] be a Lebesgue measurable space, i.e., [itex]m_{k}=m[/itex] is a Lebesgue measure. Let [itex]f: \mathbb{R^k} \to \mathbb{R}[/itex] be a [itex]m[/itex]-integrable function. Define a function [itex]\mu : \mathcal{A} \to [0,\infty][/itex] by $$ \mu(A) := \int_{A} f(x) dx $$ with [itex]A \in \mathcal{A}[/itex]. Then is [itex]\mu[/itex] a measure?

Qeustion. Is there any error in the below working? And does it need any additional hypothesis to show $mu$ is a measure? Or is it acutally not a measure?

I'm missing somewhere that [itex]f\geq 0[/itex] a.e.

The Attempt at a Solution


When [itex]A[/itex] is an empty set, the integral is obviously 0. Now suppose [itex]A_{0} , \dots , A_{n} \in \mathcal{A}[/itex] as they are disjoint. Then $$ \mu ( \bigcup_{k=0}^{n} A_{k} ) := \int_{\mathbb{R}^{k}} \chi_{\bigcup_{k=0}^{n} A_{k}}(x)f(x)dx $$ and we get $$ \chi_{\bigcup_{k=0}^{n} A_{k}} = \sum_{k=0}^{n} \chi_{A_k}$$ as [itex]A_{k}[/itex] are disjoint. Therefore, we have $$ \mu ( \bigcup_{k=0}^{n} A_{k} ) = \sum_{k=0}^{n} \mu(A_{k}). $$ Thus taking [itex]n \to \infty[/itex] we get the required result.

You have to justify why

[tex]\lim_{n\rightarrow +\infty} \mu ( \bigcup_{k=0}^{n} A_{k} )=\mu ( \bigcup_{k=0}^{+\infty} A_{k} )[/tex]
 
  • #3
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Thanks for your answer, and as you have told me I should assume that [itex]f(x) \geq 0 [/itex] a.e.. And for the last part you mentioned, I think Dominated Convergence Theorem would suffice, woudn't it? I used it in my solution..
 
  • #4
22,129
3,297
Thanks for your answer, and as you have told me I should assume that [itex]f(x) \geq 0 [/itex] a.e.. And for the last part you mentioned, I think Dominated Convergence Theorem would suffice, woudn't it? I used it in my solution..

Yes, the dominated convergence would suffice here.
 

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