I have a vague problem in a Measure and Integration book here. They ask me to consider $\mu$ the counting measure in $\mathbb{N}$ and interpret Fatou's lemma, monotone and dominated convergence theorems as statements about infinite series. I thought it would be easy but got stuck really quick…
If we consider a sequence of non negative functions $f_n\in L^1$, Fatou's lemma says that $$\int_\mathbb{N}\liminf f_n \ d\mu \leq \liminf\int_\mathbb{N}f_n \ d\mu.$$
The real problem comes when I try to understand what are this integrals. Starting with a simple function $\Phi =\sum_{i=1}^kx_i\mathcal{X}_{E_i}$, we have that $\int_\mathbb{N}\Phi \ d\mu = \sum_{i=1}^kx_i\mu(E_i) = \sum_{i=1}^kx_i|E_i|$, where $|E_i|$ is the number of elements in $E_i$. Each function $f_n$ is the $\sup$ of simple functions, but it's not clear how I should use all this together. Even if I consider a sequence $0\leq \Phi_1\leq\Phi_2\leq\ldots\leq f_n$ converging to $f_n$, they are not partial sums.
To make things worse, there is infinite $f_n$ to consider and the $\liminf$ after that. Any help is welcome to interpret all this.
Thank you.
Best Answer
If the $f_n$s are non-negative measurable functions (as in the assumption of Fatou's lemma) and $\mu_c$ is the counting measure, then $\int_Nf_nd\mu_c=\sum_{k=1}^{\infty}f_n(k)$. If the functions are $L^1(\mu_c)$ then the integral and hence the series converges to a positive real number. Fatou's lemma is then $\sum_{k=1}^{\infty}(\text{lim inf}f_n(k))\leq \text{lim inf}\sum_{k=1}^{\infty}f_n(k)$, which exists because inside the lim inf on the RHS is a squence of real numbers.
The reason why the integrals are series: Let $f$ be a non-negative function on the sigma algebra $(\mathbb{N},P(\mathbb{N}))$. Then every function on this sigma algebra is measurable, so $f$ is a non-negative measureable function. Then if $f_n=\chi_{[0,n]}f$, clearly $0\leq...\leq f_n\leq f_{n+1}\leq...$ and $f_n\to f$ pointwise. So $\text{lim}\int_N f_nd\mu_c=\int_Nfd\mu_c$ by montone convergence theorem. But what is the LHS? $\int_Nf_nd\mu_c=\int_{[0,n]}fd\mu_c=\int_0fd\mu_c+\int_1fd\mu_c+...+\int_nf\mu_c=\sum_{k=1}^nf(k)$. Then taking the limit gives you $\sum_{k=1}^{\infty}f(k)=\int_Nfd\mu_c$