Sums and integration against counting measure: Appendix 4: More details about liminf

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

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In this appendix, we fill in some details about claims made in Part IV concerning $\liminf$.

We will use the same definitions as in Part IV concerning the $\liminf$ of nets in $\Rbar$.

Now let $J$ be a (non-empty) indexing set, and suppose that we have non-negative extended real numbers $(a_j)_{j \in J}$ and $(b_j)_{j \in J}$. We claim that

\[\inf\{a_j + b_j: j \in J\}\geq \inf\{a_j:j \in J\} + \inf\{b_j:j\in J\}.\tag{$1$}\]

This is fairly standard, but here is a quick proof for those in any doubt.

Set $a=\inf\{a_j:j \in J\}$ and $b=\inf\{b_j:j\in J\}$. Then, for each $j \in J$, we have $a_j \geq a$ and $b_j \geq b$. Thus (noting that all values are in $[0,\infty]$, so there is no danger when adding) we have

\[ a_j +b_j \geq a+b\,.\]

The inequality ($1$) now follows immediately (either by the definition of infimum, or by taking the infimum over $j \in J$).

Next 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 Part IV of the main series we claimed that

\[\liminf_\alpha (y_\alpha + z_\alpha) \geq \liminf_\alpha y_\alpha + \liminf_\alpha z_\alpha\,.\tag{$2$}\]

To see this, for each $\alpha \in A$, set

\[s_\alpha = \inf \{y_\beta :\beta \in A, \beta \geq \alpha\} \in [0,\infty]\,,\] and

\[t_\alpha= \inf \{z_\beta :\beta \in A, \beta \geq \alpha\} \in [0,\infty]\,.\]

Also set

\[y=\liminf_\alpha y_\alpha = \sup_{\alpha\in A} s_\alpha = \lim_\alpha s_\alpha\,,\] and

\[z=\liminf_\alpha z_\alpha = \sup_{\alpha\in A} t_\alpha = \lim_\alpha t_\alpha\,.\]

Note that, for each $\alpha \in A$, by ($1$),

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

Taking the limit over $\alpha$ of both sides gives us $(2)$.

It would be nice to avoid using the algebra of limits here by taking the supremum of both sides instead. Unfortunately the standard inequality for the supremum of sums is the wrong way round! However, the nets $s_\alpha$ and $t_\alpha$ are monotone increasing here, so it is actually true in this setting that

\[\sup_\alpha(s_\alpha + t_\alpha) = \sup_\alpha s_\alpha + \sup_\alpha t_\alpha\,.\]

So (if you check the details of that claim) you don't have to appeal to the algebra of limits.

Now let $X$ be a set with exactly two elements, say $X=\{x_1,x_2\}$, and let $(f_\alpha)_{\alpha \in A}$ be a net of functions from $X$ to $[0,\infty]$. Taking $y_\alpha=f_\alpha(x_1)$ and $z_\alpha=f_\alpha(x_2)$, ($2$) gives us

\[\liminf_\alpha (f_\alpha(x_1) + f_\alpha(x_2)) \geq \liminf_\alpha f_\alpha(x_1) + \liminf_\alpha f_\alpha(x_2)\,,\] which we can rewrite as

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

That gives us Fatou's Lemma for nets and sums in the case where $X$ has two points. We'll repeat that in the next appendix, as part of our proof for general sets.