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.

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    $\begingroup$ You don't need to go to use simple functions. Just write the integrals in terms of $f_n$. For Fatou's lemma, if $f:=\liminf_n f_n$, then $\sum_{k=1}^{\infty}f(k)\leq\liminf_n \sum_{k=1}^{\infty}f_n(k)$. All we are using is that $\int_{\mathbb{N}}g(x)d\mu=\sum_{k=1}^{\infty}g(k)$ when $\mu$ is the counting measure on $\mathbb{n}$. Maybe you want to prove that too!? Then use the simple functions definition of the integral to prove it. Afterwards it is just doing a translation from the language of integrals to the series. $\endgroup$ – Pp.. Jan 19 '15 at 21:29
  • $\begingroup$ But how the integrals become series? Maybe that is my problem! $\endgroup$ – Integral Jan 19 '15 at 21:33
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    $\begingroup$ Just take the simple functions $\sum_{k=1}^{\infty} f(k)\chi_{\{k\}}+\text{ any simple function on sets that don't contain natural numbers}$. $\endgroup$ – Pp.. Jan 19 '15 at 22:11
  • $\begingroup$ I understand the idea, but any simple function must be on sets containing natural numbers, after all, the $\sigma$-algebra here is $\mathcal{P}(\mathbb{N})$. Thanks. $\endgroup$ – Integral Jan 19 '15 at 22:36
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    $\begingroup$ Ah! I thought it was on the reals. Then it is even easier. It is only the first summation I wrote. $\endgroup$ – Pp.. Jan 20 '15 at 1:15

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 $\sum_{k=1}^{\infty}f(k)=\int_Nfd\mu_c$


By writting $\mu=\sum\limits_{n\in\mathbb{N}}\delta_n$, where $\delta_n$ is the Dirac measure at $n$, we can show that $\int_Nh d\mu=\sum\limits_nh(n)$ for any positive function $h$ (or any integrable function $h$)


I would like to provide a clearer proof to the post. As a start, we recall Fatou's lemma, that is, $$\int_{\Omega}\liminf\:h \:d\mu \leq \liminf \int_{\Omega}h\:d\mu $$

We first show the sketch of the proof:

$\textbf{First}$, we show that for any $h$: $(\mathbb{N},P(\mathbb{N})) \rightarrow (\mathbb{R}^{+},B(\mathbb{R}^+))$ (i.e. nonnegative measurable function), we have \begin{equation} \sum_{k=1}^{\infty}h(k)=\int_{\mathbb{N}}h(k)\:d\mu_{c}(k), \tag{1} \end{equation}

where $\mu_{c}$ is a counting measure, $\mu_c$ is defined on $(\mathbb{N},P(\mathbb{N}))$, where $P(\mathbb{N})$ is the powerset of natural numbers.

$\textbf{Second}$, let us define $f(k) \equiv \liminf_{n\rightarrow \infty}\:f_n(k) \;\forall k\in\mathbb{N} $. Since this newly defined $f$ is such that $f$: $\mathbb{N} \rightarrow \mathbb{R}^+$ and is measurable, then from (1), $$\sum_{k=1}^{\infty} \liminf_{n\rightarrow\infty}f_n(k) = \int_{\mathbb{N}}\liminf f_n(k)\;d\mu_c \leq \liminf \int_{\mathbb{N}}f_n \;d\mu_c = \liminf \sum_k f_n (k),$$ where the inequality follows from fatou's lemma above, and the last equality follows from (1) above since $f_n$: $\mathbb{N} \rightarrow \mathbb{R}^+$ and is measurable. $\textbf{Our job is therefore reduced to proving (1)}$

$\textbf{Proof of (1)}$:

For any nonnegative measurable function $h$: $(\mathbb{N},P(\mathbb{N})) \rightarrow (\mathbb{R}^{+},B(\mathbb{R}^+))$, we define $h_n \equiv \chi_{[1,n]} h$. Note first that $h_n \rightarrow h$ as $n\rightarrow \infty$. In addition, $h_n$ forms a sequence of nonnegative increasing function.

\begin{align} \int_{\mathbb{N}} h_n \; d\mu_c = \int_{[1,n]} h\; d\mu_c &= \int_1 h\; d\mu_c + \int_2 h\; d\mu_c + ... + \int_n h\;d\mu_c \\ &= h(1)+h(2)+...+h(n)=\sum_{k=1}^{n} h(k)\\ \end{align}

Then $\lim_{n\rightarrow\infty} \int_{\mathbb{N}}h_n\;d\mu_c = \lim_{n\rightarrow\infty} \sum_{k=1}^{n} h(k) = \sum_{k=1}^{\infty} h(k) \tag{2}$

However, by $\textbf{monotone convergence theorem}$, we know that $$\lim_{n\rightarrow\infty} \int_{\mathbb{N}}h_n\;d\mu_c = \int_{\mathbb{N}} h\;d\mu_c\tag{3},$$ since $0\leq h_n \uparrow h$. Equating (2) and (3) yields (1)

$ \square$


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