Countable additivity of the Lebesgue integral for union of a. e. disjoint sets

Question

Let $f \in \mathcal{L}^1(X)$ and $X = \cup_{n \in \mathbb{N}} A_n$ with measurable sets $A_n$ such that $\mu(A_i \cap A_j) = 0$ if $i \neq j$. Show that $$ \int_X f \mathrm{d} \mu = \sum_{n \in \mathbb{N}} \int_{A_n} f \mathrm{d} \mu. $$ This is a homework question that I'm not able to solve. It has been discussed in other threads (and I understand the ideas there) for the case that the $A_i$ are pairwise disjoint (and not just disjoint almost everywhere). Maybe someone can provide me with a hint in right direction. It is obvious that this question wants me to apply monotone/ dominating convergence theorem but I just can't figure out how.

Thank you in advance.

Answer 1

I will try and get you started.

As you mentioned, you should use one of the convergence theorems. So let's construct a sequence of functions so we may apply it. First, let $f_n=f\chi_{E_n}$ where $E_n=\bigcup_{k=1}^n A_k$.

We see that $f_n \rightarrow f$ pointwise as $n \rightarrow \infty$

We also have that every term of the sequence has absolute value bounded above by an $L^1$ function.

i.e. $|f_n| \leq |f|$ with $|f| \in L^1$

So we realize that we meet the conditions of the dominated convergence theorem.

$$\lim_{n \rightarrow \infty} \int f_n d\mu=\int f d\mu$$

$$\lim_{n \rightarrow \infty} \int f \chi_{\cup_n A_n}d\mu=\lim_{n \rightarrow \infty} \int f (\sum_k^n \chi_{A_k}-\sum_{i,j i \neq j}^n \chi_{A_i \cap A_j})d \mu $$

Try and go from here. Good luck

Answer 2

Lemma: If $(A_n)_{n=1}^\infty$ is disjoint almost everywhere (i.e. $\mu(A_i \cap A_j) = 0$ for all $i \neq j$). Then $\sum_n\mu(A_n \cap B) = \mu(B)$.

Proof: Let $G:= \bigcup_{i \neq j}A_i \cap A_j$. Then $\mu(G) = 0$ (here it is important that our union is indexed by a countable collection). Thus $$\sum_n \mu(A_n \cap B) = \sum_n \mu(A_n \cap B \cap G^c) = \mu\left(\bigcup_n A_n \cap B \cap G^c\right) = \mu(B \cap G^c) = \mu(B) $$

and the lemma is proven. $\quad \square$

Let $f= I_B$ with $B$ a Borelset. Then

$$\sum_n\int_{A_n} f d \mu = \sum_n \int_{A_n} I_B d \mu = \sum_n \int_{A_n\cap B}d\mu = \sum_n \mu(A_n \cap B) = \mu(B) =\int_X f d \mu $$

and your theorem holds for indicator functions. From here the proof is the same as the other proofs. You prove it for sums of indicators, then lift it to positive functions using monotone convergence theorem and then prove it for general functions by writing such a function as a difference of two positive functions.

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Interesting addendum: The statement you provided also holds for semi-integrable functions.