Sums and integration against counting measure: Appendix 5: Fatou's Lemma for sums without using Measure and Integration

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For other posts in this series, see

https://explaining-maths.blogspot.com/search?q=Sums+and+integration+against+counting+measure

In case the MathJax mathematics below does not display properly in your browser, I will make a PDF of the main article available via my WordPress blog

In this appendix, following on from Appendices 3 and 4, we see how the argument from Part IV of the main series gives us Fatou's Lemma for sums without using Measure and Integration. As in Appendix 3, we will assume that we know standard facts about finite sums. We will use the same definitions as in Part IV concerning the $\liminf$ of nets in $\Rbar$ and of nets of functions from $X$ to $\Rbar$.

We base our arguments on the following facts about sums (see earlier posts).

Let $X$ be a set, and let $f$ be a function from $X$ to $[0,\infty]$. Then
\[\sum_{x \in X} f(x) = \sup \left\{ \sum_{x \in E} f(x): E \textrm{ is a finite subset of } X \right\}.\tag{$1$}\]
It follows that, whenever $B_1$ and $B_2$ are subsets of $X$ with $B_1\subseteq B_2$, we have

\[\sum_{x \in B_1} f(x) \leq \sum_{x \in B_2} f(x)\,.\tag{$2$}\]

Also, if $f$ and $g$ are functions from $X$ to $[0,\infty]$ with $f(x)\leq g(x)$ for all $x\in X$, then we can write $0 \leq f \leq g$, and we then have

\[\sum_{x \in X} f(x)\leq \sum_{x \in X} g(x)\,.\tag{$3$}\]
(But maybe we don't need that one.)

Our final preliminary facts are about $\liminf$ and nets. If you don't know about nets, you could pretend that they are sequences if you like. But then you might miss the fact that the usual version of Fatou's Lemma is for sequences and not for nets. Fatou's Lemma for sums is different, because it works for nets too.

Suppose that $(y_\alpha)_{\alpha \in A}$ and $(z_\alpha)_{\alpha \in A}$ are both nets in $[0,\infty]$ (where $A$ is a directed set). In Appendix 4 we proved that (as claimed in Part IV of the main series)

\[\liminf_\alpha (y_\alpha + z_\alpha) \geq \liminf_\alpha y_\alpha + \liminf_\alpha z_\alpha\,.\tag{$4$}\]

Let's rewrite that to look a bit more like Fatou's Lemma:

\[ \liminf_\alpha y_\alpha + \liminf_\alpha z_\alpha \leq \liminf_\alpha (y_\alpha + z_\alpha) \,.\tag{$4'$}\]

Fatou's Lemma for sums and nets

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

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

Proof

We prove the result first when $X$ has at most two points, then for finite sets $X$ and finally for general sets $X$.

If $X$ is empty, then both sides of ($*$) are $0$, so equality holds.

If $X$ has just one point, say $x_1$, then both sides of ($*$) are equal to $\liminf_\alpha f_\alpha(x_1)$, and again equality holds.

Now suppose that $X$ has exactly two points, say $X=\{x_1,x_2\}$ with $x_1\neq x_2$. As we saw in Appendix 4, taking $y_\alpha=f_\alpha(x_1)$ and $z_\alpha=f_\alpha(x_2)$, ($4'$) gives us

\[\liminf_\alpha f_\alpha(x_1) + \liminf_\alpha f_\alpha(x_2) \leq \liminf_\alpha (f_\alpha(x_1) + f_\alpha(x_2))\,,\] which is exactly what ($*$) says in this case.

We can now prove the result for finite sets $X$ by induction on the number of points in $X$. We know the result holds for sets with at most two points. Now suppose that $X$ has $n$ points for some $n>2$ and that the result holds for all sets with at most $n-1$ points. 

We have $X=\{x_1,x_2,\dots,x_{n-1},x_n\,\}$, say, where the $x_i$ are distinct. Set $Y=\{x_1,x_2,\dots,x_{n-1}\,\}$ and, for each $\alpha \in A$, set 

\[y_\alpha=\sum_{x \in Y} f_\alpha(x) = \sum_{k=1}^{n-1} f_\alpha(x_k)\quad \textrm{and}\quad z_\alpha=f_\alpha(x_n)\,.\]

Then  we have

\[y_\alpha + z_\alpha = \sum_{x\in X} f_\alpha(x)\,.\]

By ($4'$) we have

\[\liminf_\alpha y_\alpha + \liminf_\alpha z_\alpha \leq \liminf_\alpha(y_\alpha+ z_\alpha) = \liminf_\alpha \sum_{x\in X} f_\alpha(x)\,.\tag{$5$}\]

By the inductive hypothesis (applied to $Y$) we have

\[\sum_{x\in Y}\liminf_\alpha f_\alpha(x) \leq \liminf_\alpha \sum_{x\in Y} f_\alpha(x) = \liminf_\alpha y_\alpha\,.\tag{$6$}\]

Finally we combine ($5$) and ($6$):

\[\begin{aligned}\sum_{x \in X} \liminf_\alpha f_\alpha(x) &= \sum_{x\in Y}\liminf_\alpha f_\alpha(x) + \liminf_\alpha f_\alpha(x_n)\\
&=\sum_{x\in Y}\liminf_\alpha f_\alpha(x) + \liminf_\alpha z_\alpha\\
&\leq \liminf_\alpha y_\alpha +\liminf_\alpha z_\alpha \leq \liminf_\alpha \sum_{x\in X} f_\alpha(x)\,.\end{aligned}\]

This proves the result when $X$ has $n$ points. The induction may proceed, and the result holds whenever the set $X$ is finite.

Note: We didn't really need to prove the case $n=2$ separately here, as it could have been proved as part of the main induction (as long as we have ($4'$) available). However I felt it was worth seeing that the case $n=2$ is essentially just a restatement of $(4')$.

Finally, as in Part IV, we prove the general case using the result for finite sets. This is essentially a repeat of what we did there, but written entirely in terms of sums.

For convenience, here is what we are trying to prove again.

Let $X$ be a set (which may be uncountably infinite), and let $(f_\alpha)_{\alpha\in A}$ be a net of functions from $X$ to $[0,\infty]$ (as above). We wish to prove that

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

To prove this, first set $\displaystyle L = \liminf_\alpha \sum_{x\in X} f_\alpha(x)$ (the right hand side of $(*)$), and set $f=\liminf_\alpha f_\alpha$ (i.e., the function $x \mapsto \liminf_\alpha f_\alpha(x)$). Then we can rewrite our target $(*)$ as 

\[\sum_{x \in X} f (x) \leq L\,.\tag{$**$}\]

We have already proved the result for finite sets, so for every finite set $K\subseteq X$, we have

\[\sum_{x \in K} f(x) \leq \liminf_\alpha \sum_{x \in K} f_\alpha(x)\,,\] and of course (using ($2$))

\[\liminf_\alpha \sum_{x \in K} f_\alpha(x) \leq \liminf_\alpha \sum_{x \in X} f_\alpha(x)=L. \]

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

\[\sum_{x \in K} f(x) \leq L\,.\]

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

\[\sum_{x \in X} f (x) \leq L\,,\] as required. $\quad\square$

I should really give lots of examples and some applications of Fatou's Lemma for sums and nets. (I have used this a lot when working with things like ordinal sequences of elements of Abstract Swiss Cheese space.) That is something for future posts!

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