Consider the measure space $(\mathbb{R}^2,\mathcal{B}(\mathbb{R^2}),\lambda^2)$ and let $f=\sum_{n=1}^\infty f_n$ where
$$f_n=\frac{1}{\sqrt{n}}1_{B_n}$$ and $B_n=\left [-1/n^2,1/n^2 \right )\times \left [ -\sqrt{n},\sqrt{n} \right ), n\geq 1$.
- Determine $\int f\; d\lambda^2$
I have shown that the function $f$ is an element in $\mathcal{M}_{\bar{\mathbb{R}}}^+(\mathcal{B}(\mathbb{R}^2))$ but I am unsure on how to continue from this point (knowing that it can be determined from it being an element in the given set). How do I determine such a integral from this point?
(Assuming $\lambda^2$ is the 2-dimensional Lebesgue measure.)
The $f_n$ are positive functions, therefore $g_N := \sum_{n = 1}^{N}{f_n}$ are positive functions, increasing in n. Hence the Monotone Convergence Theorem yields $$ \int{f(x) \, \, \lambda^2(dx)} = \int{\lim_{N \to \infty} g_N(x) \, \, \lambda^2(dx)} = \lim_{N \to \infty} \int{ g_N(x) \, \, \lambda^2(dx)}. $$ For each N we have $$ \int{ g_N(x) \, \, \lambda^2(dx)} = \sum_{n = 1}^{N} { \int{ f_n(x) \, \, \lambda^2(dx)} } $$ using the linearity of the integral. As each $f_n$ is just a scaled indicator function we further get $$ \int{ f_n(x) \, \, \lambda^2(dx)} = \frac{1}{\sqrt{n}} \, \lambda^2(B_n) = \frac{1}{\sqrt{n}} \lambda^2( [−1/n^2,1/n^2) \times [−\sqrt{n},\sqrt{n}) )\\ = \frac{1}{\sqrt{n}} \, \lambda([−1/n^2,1/n^2) ) \, \lambda([−\sqrt{n},\sqrt{n})) \\ = \frac{2(1/n^2) \cdot 2\sqrt{n} }{\sqrt{n}} = \frac{4}{n^2}. $$
Summing up: $$ \int{f(x) \, \, \lambda^2(dx)} = \lim_{N \to \infty} \sum_{n = 1}^{N} { \int{ f_n(x) \, \, \lambda^2(dx)} } = 4 \sum_{n = 1}^{\infty}{\frac{1}{n^2}}. $$
For the value of this series see e.g. https://en.wikipedia.org/wiki/Basel_problem .