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?
Working. 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.
Question. Is there any error in the above working? And does it need any additional hypothesis to show $\mu$ is a measure? Or is it actually not a measure?