Prove that if \(f, g\in L^+(\RR)\) then \[ \int(f +g) = \int f + \int g .\] Extend this to establish that if \(\ts{ f_k} \subseteq L^+(\RR^n)\) then \[ \int \sum_k f_k = \sum_k \int f_k .\]
Let \(\ts{E_j}_{j\in \NN} \subseteq \mathcal{M}(\RR^n)\) with \(E_j \nearrow E\). Use the countable additivity of \(\mu_f\) on \(\mathcal{M}(\RR^n)\) established above to show that \[ \mu_f(E) = \lim_{j\to \infty } \mu_f(E_j) .\]
Solution
Here \(L^+\) is the set of measurable functions with values in \([0,\infty]\), and \(\mu_f(A) = \int_A f\) for \(f \in L^+\).
1. For \(f, g \in L^+\), \(\int (f + g) = \int f + \int g\).
1.1. The identity holds for nonnegative simple \(s = \sum_i a_i \chi_{A_i}\) and \(t = \sum_j b_j \chi_{B_j}\), where \((A_i)\) and \((B_j)\) are finite measurable partitions of the domain.
\(s + t = \sum_{i,j}(a_i + b_j)\chi_{A_i \cap B_j}\), so \(\int (s+t) = \sum_{i,j}(a_i + b_j)\,m(A_i \cap B_j)\). Summing over \(j\) first in the \(a_i\) terms and over \(i\) first in the \(b_j\) terms gives \(\sum_i a_i m(A_i) + \sum_j b_j m(B_j)\).
1.2. Q.E.D.
Choose simple \(0 \le s_k \uparrow f\) and \(0 \le t_k \uparrow g\). Then \(s_k + t_k \uparrow f + g\), and the monotone convergence theorem with step 1 gives \(\int(f+g) = \lim_k (\int s_k + \int t_k) = \int f + \int g\).
2. For \(f_k \in L^+\), \(\int \sum_k f_k = \sum_k \int f_k\).
By step 1 and induction, \(\int\sum_{k=1}^N f_k = \sum_{k=1}^N\int f_k\). The partial sums increase to \(\sum_k f_k\), so the monotone convergence theorem gives the claim.
3. If \(E_j \nearrow E\), then \(\mu_f(E_j) \to \mu_f(E)\).
Put \(E_0 = \emptyset\) and \(D_k = E_k \setminus E_{k-1}\). The \(D_k\) are disjoint with \(\bigcup_k D_k = E\) and \(\bigcup_{k \le j} D_k = E_j\). Step 2 applied to \(f\chi_{D_k}\) gives countable additivity of \(\mu_f\): \(\mu_f(E) = \sum_k \mu_f(D_k)\). Step 1 gives \(\mu_f(E_j) = \sum_{k \le j}\mu_f(D_k)\), the \(j\)th partial sum, which converges to \(\mu_f(E)\).