Sums and integration against counting measure: Part IV

For other posts in this series, see
https://explaining-maths.blogspot.com/search?q=Sums+and+integration+against+counting+measure

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This post is a little more advanced, in as much as it assumes some knowledge of nets rather than just sequences.

In this post we will look at the three big convergence theorems for Lebesgue integration, concerning sequences of functions, and see how these go wrong if you use nets instead. Then we will see that they go right again (for nets) if we work with the special case of counting measure.

Let's jump straight in with an example of a 'bad' net of functions when working with Lebesgue measure $\lambda$ on $[0,1]$. Our directed set will be the set $\mathcal{F}$ of all finite subsets of $[0,1]$, partially ordered by inclusion. Our net of functions $g_E\jq(E\in\mathcal{F})$ will just be the characteristic functions of these sets, so $g_E=\chi_E\jq(E\in\mathcal{F})$. It is easy to see that this is a monotone increasing net of non-negative measurable functions which converges pointwise on $[0,1]$ to the constant function $g\equiv 1$. This measurable function $g$ also 'dominates' all of the functions $g_E$ in the sense of the Dominated Convergence Theorem. However we have

\[\int_{[0,1]} g_E\,\rd\lambda = 0\jq (E\in\mathcal{F})\,,\] while

\[\int_{[0,1]} g\,\rd\lambda = 0\,.\]

So the integrals fail to converge to the integral of the limit (showing that the Monotone Convergence Theorem and Dominated Convergence Theorem both fail in this setting). Of course $g$ is also the $\liminf$ of this net of functions, and so Fatou's Lemma also fails here.

What seems to be less well known is that the results work again if you work with counting measure. 

Let $X$ be a set. Let's work with the whole power set of $X$ (so that all functions are measurable), and take our measure $\mu$ to be counting measure on $X$. Let's jump straight to Fatou's Lemma for nets, because everything follows once you have Fatou.

We will need to work with $\liminf$ and $\limsup$ of nets of extended real numbers here, so let's recall how that works.

We denote by $\Rbar$ the extended real line $[-\infty,\infty]$.

Let $(y_\alpha)_{\alpha \in A}$ be a net in $\Rbar$, where $A$ is a directed set. For each $\alpha\in A$, set 

\[s_\alpha= \inf\{y_\beta:\beta \in A, \beta \geq \alpha\,\}\,\in\Rbar.\]

As $\alpha$ increases, these sets decrease (not necessarily strictly) and so the $s_\alpha$ are nondecreasing as $\alpha$ increases. We then define

\[\liminf_\alpha y_\alpha = \sup_{\alpha \in A} s_\alpha\, \in \Rbar.\]

Similarly we define \[S_\alpha= \sup\{y_\beta:\beta \in A, \beta \geq \alpha\,\}\,\in \Rbar\] and

\[\limsup_\alpha y_\alpha = \inf_{\alpha \in A} S_\alpha\, \in \Rbar.\]

All of the usual relationships between $\liminf$, $\limsup$ and $\lim$ apply for nets in $\Rbar$. In particular, with the above notation, we also have

\[\liminf_\alpha y_\alpha = \lim_\alpha s_\alpha\] and

\[\limsup_\alpha y_\alpha=\lim_\alpha S_\alpha\,.\]

Next suppose that $(y_\alpha)_{\alpha \in A}$ and $(z_\alpha)_{\alpha \in A}$ (with $A$ as above) are both nets in $[0,\infty]$. Then you can easily check that

\[\limsup_\alpha (y_\alpha + z_\alpha) \leq \limsup_\alpha y_\alpha + \limsup_\alpha z_\alpha\,,\] and (more importantly for us)

\[\liminf_\alpha (y_\alpha + z_\alpha) \geq \liminf_\alpha y_\alpha + \liminf_\alpha z_\alpha\,.\tag{$*$}\]

Now let  $(f_\alpha)_{\alpha \in A}$ be a net of functions from $X$ to $\Rbar$. Then we define the functions $\liminf_\alpha f_\alpha$ and $\limsup_\alpha f_\alpha$ from $X$ to $\Rbar$ pointwise as usual, giving us the functions $x \mapsto \liminf_\alpha f_\alpha(x)$ and $x \mapsto \limsup_\alpha f_\alpha(x)$ respectively.

Recall that we are working with counting measure $\mu$ on $X$. We wish to establish the following version of Fatou's Lemma. 

Fatou's Lemma for nets when working with counting measure

Let  $(f_\alpha)_{\alpha \in A}$ be a net of functions from $X$ to $[0,\infty]$. Then

\[\int_X \liminf_\alpha f_\alpha \, \rd \mu \leq \liminf_\alpha \int_X f_\alpha \,\rd\mu\,.\tag{$1$}\]

Note that, in terms of sums over sets (as discussed in earlier posts), $(1)$ just says that

\[\sum_{x\in X} \liminf_\alpha f_\alpha(x) \leq \liminf_\alpha \sum_{x \in X} f_\alpha(x)\,.\tag{$2$}\]

Moreover, in the case where the set $X$ is finite we can prove $(2)$ (and hence $(1)$) by a simple induction using $(*)$. Indeed, when $X$ has just one point (or no points) the result is trivial, and when $X$ has two points, $(2)$ is essentially a restatement of $(*)$. I'll leave the rest of the induction as an exercise (but can provide details on request).

So, let's prove $(1)$ for a general set $X$, which may be uncountably infinite. Set $\displaystyle L = \liminf_\alpha \int_X f_\alpha \,\rd\mu$ (the right hand side of $(1)$), and set $f=\liminf_\alpha f_\alpha$. Then we can rewrite our target $(1)$ as 

\[\int_X f \, \rd \mu \leq L\,.\tag{$1'$}\]

By our comments on finite sets above, we note that for every finite set $K\subseteq X$, we have

\[\int_K f \, \rd \mu \leq \liminf_\alpha \int_K f_\alpha \,\rd\mu\,,\] and of course

\[\liminf_\alpha \int_K f_\alpha \,\rd\mu \leq \liminf_\alpha \int_X f_\alpha \,\rd\mu\,=L. \]

So, for every finite subset $K$ of $X$, we have

\[\int_K f \, \rd \mu \leq L\,.\]

Taking $\sup$ over all finite sets $K\subseteq X$ gives us $(1')$,

\[\int_X f \, \rd \mu \leq L\,,\] as required. $\quad\square$

The last part is justified using a claim (unproved) from Part I of this series of posts, where we noted (in terms of sums) that

\[\displaystyle \sum_{x \in X} f(x) = \sup \left\{ \sum_{x \in E} f(x): E \textrm{ is a finite subset of } X \right\}.\]

Once we have Fatou's Lemma in this setting, it is easy to deduce the Montone Convergence Theorem (and the slightly more general Convergence from Below Theorem if you want it), and the Dominated Convergence Theorem (this can be deduced from Fatou in the usual way).

I'll probably write some more posts where I give some of the missing details.