Let $ (X,\mathcal{M}, \mu) $ be a measure space such that $\mu(X) < \infty$ and let $f,g,h \in L^1(\mu)$. For $n \in \mathbb{N}$, define
$$B_n = \{x \in X: |f(x)| +|g(x)| \leq n \} \in \mathcal{M}$$
Prove that
$$\lim_{n \rightarrow \infty} \int _ {B_n} h d\mu = \int_X hd\mu$$
This is my question.
Since $\mu (X) < \infty$ and $f,g \in L^1(\mu)$, I show that the $f,g < \infty$ a.e.
I thought that I need to show that $B_n \rightarrow X$ and $B_n$ is an increasing set.
And maybe If I define $\phi(E) :=\int_{E}h d\mu$ for $E \in \mathcal{M}$ and show that $\phi$ is a measure, I could prove the result by countable additivity.
But I'm not sure about how to prove it.
- $\mu$ is a positive measure.
First, it is clear from the definition that $B_n \subset B_{n+1}$. Therefore, we only need to show that : $$\mu\left(\bigcup_{n \geq 0} B_n\right) = \mu(X) \tag{1}$$
To do this, we see that : \begin{align} X - \bigcup_{n \geq 0} B_n &= \bigcap_{n\geq 0} (X-B_n)\\ &\subset |f|^{-1}(\{+\infty\}) \cup |g|^{-1}(\{+\infty\}) \end{align} Therefore, because $|f|,|g|<\infty$ almost everywhere, we know that : $$\mu\left(X - \bigcup_{n \geq 0} B_n\right) = 0$$ Since $\mu(X) < \infty$, we have proved $(1)$.
Then, as $h$ is $L^1$, by dominated convergence, we have : $$\lim_{n\to +\infty}\int_{B_n} h \text d\mu= \int_{\bigcup_{n\geq 0}B_n} h \text d\mu= \int_x h\text d\mu$$