Let \(f\) be integrable over a measurable set \(A\) and suppose \[ A=\bigcup_{n=1}^\infty A_n, \] where the \(A_n\) are pairwise disjoint. Show that \[ \int_A f=\sum_{n=1}^\infty\int_{A_n}f \] and that the series on the right is absolutely convergent.
Let \(\mu\) be Lebesgue measure on \(\mathbb R^2\) and let \(f\in L^1(\mathbb R^2)\). Show that \[ \lambda(E):=\int_E f\,d\mu \] defines a finite signed Borel measure on \(\mathbb R^2\). In particular, if \(f\ge0\) almost everywhere, then \(\lambda\) is a finite positive Borel measure and \(d\lambda=f\,d\mu\).
For \(f(x,y)=x^2+y^2\) and \(D\) the unit disc, compute \(\lambda(D)\).
Solution
1. The integral is countably additive over the disjoint partition.
Let \[ B_N=\bigcup_{n=1}^N A_n. \] Then \(\mathbf 1_{B_N}f\to \mathbf 1_Af\) pointwise and \[ |\mathbf 1_{B_N}f|\le \mathbf 1_A|f|\in L^1. \] By dominated convergence, \[ \int_A f =\lim_{N\to\infty}\int_{B_N}f =\lim_{N\to\infty}\sum_{n=1}^N\int_{A_n}f =\sum_{n=1}^\infty\int_{A_n}f. \] Moreover, \[ \sum_{n=1}^\infty\left|\int_{A_n}f\right| \le \sum_{n=1}^\infty\int_{A_n}|f| =\int_A|f|<\infty. \] Hence the series is absolutely convergent.
2. The density defines a finite signed Borel measure.
For every Borel set \(E\subseteq\mathbb R^2\), define \[ \lambda(E)=\int_E f\,d\mu. \] Clearly \(\lambda(\varnothing)=0\). If \((E_n)\) are pairwise disjoint Borel sets, then applying 1 to \(f\) on \(\bigcup_nE_n\) gives \[ \lambda\!\left(\bigcup_{n=1}^\infty E_n\right) =\sum_{n=1}^\infty\lambda(E_n), \] and the series is absolutely convergent. Thus \(\lambda\) is countably additive.
It is finite in total variation because \[ |\lambda(E)|\le\int_E|f|\,d\mu\le\|f\|_1, \] and in fact \(|\lambda|(\mathbb R^2)=\int_{\mathbb R^2}|f|\,d\mu<\infty\). Therefore \(\lambda\) is a finite signed Borel measure. If \(f\ge0\) almost everywhere, then \(\lambda(E)\ge0\) for every Borel \(E\), so \(\lambda\) is a finite positive Borel measure.
3. Compute the mass of the unit disc.
For \(f(x,y)=x^2+y^2\), the density is nonnegative, so \(\lambda\) is a positive measure. In polar coordinates, \[ \lambda(D) =\int_D(x^2+y^2)\,dx\,dy =\int_0^{2\pi}\int_0^1 r^2\,r\,dr\,d\theta. \] Hence \[ \lambda(D) =2\pi\left[\frac{r^4}{4}\right]_0^1 =\frac{\pi}{2}. \]