A measure related with Reimann integral

[julypraise]

Homework Statement

Let $( \mathbb{R}^k , \mathcal{A} , m_{k} )$ be a Lebesgue measurable space, i.e., $m_{k}=m$ is a Lebesgue measure. Let $f: \mathbb{R^k} \to \mathbb{R}$ be a $m$-integrable function. Define a function $\mu : \mathcal{A} \to [0,\infty]$ by $$ \mu(A) := \int_{A} f(x) dx $$ with $A \in \mathcal{A}$. Then is $\mu$ 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 $A$ is an empty set, the integral is obviously 0. Now suppose $A_{0} , \dots , A_{n} \in \mathcal{A}$ 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 $A_{k}$ are disjoint. Therefore, we have $$ \mu ( \bigcup_{k=0}^{n} A_{k} ) = \sum_{k=0}^{n} \mu(A_{k}). $$ Thus taking $n \to \infty$ we get the required result.

 

[micromass]

Homework Statement

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

The Attempt at a Solution

When $A$ is an empty set, the integral is obviously 0. Now suppose $A_{0} , \dots , A_{n} \in \mathcal{A}$ 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 $A_{k}$ are disjoint. Therefore, we have $$ \mu ( \bigcup_{k=0}^{n} A_{k} ) = \sum_{k=0}^{n} \mu(A_{k}). $$ Thus taking $n \to \infty$ we get the required result.

You have to justify why

$$\lim_{n\rightarrow +\infty} \mu ( \bigcup_{k=0}^{n} A_{k} )=\mu ( \bigcup_{k=0}^{+\infty} A_{k} )$$

 

[julypraise]

Thanks for your answer, and as you have told me I should assume that $f(x) \geq 0$ 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..

 

[micromass]

Thanks for your answer, and as you have told me I should assume that $f(x) \geq 0$ 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.