Part 6: Showing the basic four infradistribution properties: Normalization, monotonicity, concavity, and Lipschitzness (specifically 1-Lipschitzness). For normalization, observe that regardless of $n$,

$K_{:n}\left(x_0\right)(\lambda x_{1:n+1}.{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i}1)=K_{:n}\left(x_0\right)\left(1\right)=1$

$K_{:n}\left(x_0\right)(\lambda x_{1:n+1}.{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i}0)=K_{:n}\left(x_0\right)\left(0\right)=0$

The limit of the all-1 sequence is 1 and the limit of the all-0 sequence is 0, so:

$K_{:\infty }\left(x_0\right)\left(1\right)=1$

$K_{:\infty }\left(x_0\right)\left(0\right)=0$

For monotonicity, let $f^{\prime }\ge f.$ Then, regardless of $n$,

$K_{:n}\left(x_0\right)(\lambda x_{1:n+1}.{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i}{f}^{\prime }(x_{1:n+1},x_{n+2:\infty }\left)\right)$

$\ge K_{:n}\left(x_0\right)(\lambda x_{1:n+1}.{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i}f(x_{1:n+1},x_{n+2:\infty }\left)\right)$

This is because, since $f^{\prime }\ge f$, the "extend with worst-case outputs" function is bigger for $f^{\prime }$ than $f$, and then by monotonicity for $K_{:n}\left(x_0\right)$, the inequality transfers to the outside.

Accordingly,

$K_{:\infty }\left(x_0\right)\left(f^{\prime }\right)={lim}_{n\to \infty }{K}_{:n}\left(x_0\right)(\lambda x_{1:n+1}.{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i}{f}^{\prime }(x_{1:n+1},x_{n+2:\infty }\left)\right)$

$\ge {lim}_{n\to \infty }{K}_{:n}\left(x_0\right)(\lambda x_{1:n+1}.{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i}f(x_{1:n+1},x_{n+2:\infty }\left)\right)=K_{:\infty }\left(x_0\right)\left(f\right)$

Now for concavity. For all $n$, 

$K_{:n}\left(x_0\right)(\lambda x_{1:n+1}.{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i}(pf+(1-p)f^{\prime }\left)\right(x_{1:n+1},x_{n+2:\infty }\left)\right)$

$\ge K_{:n}\left(x_0\right)(\lambda x_{1:n+1}.p{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i}f(x_{1:n+1},x_{n+2:\infty })$

$+(1-p){inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i}{f}^{\prime }(x_{1:n+1},x_{n+2:\infty }))$

$\ge p{K}_{:n}\left(x_0\right)(\lambda x_{1:n+1}.{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i}f(x_{1:n+1},x_{n+2:\infty }\left)\right)$

$+(1-p)K_{:n}\left(x_0\right)(\lambda x_{1:n+1}.{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i}{f}^{\prime }(x_{1:n+1},x_{n+2:\infty }\left)\right)$

This was by monotonicity (minimizing over the two parts separately produces lower values, and monotonicity transfers that to the outside), and concavity respectively, because $K_{:n}\left(x_0\right)$ is an infradistribution. And now:

$K_{:\infty }\left(x_0\right)(pf+(1-p\left)f^{\prime }\right)$

$={lim}_{n\to \infty }{K}_{:n}\left(x_0\right)(\lambda x_{1:n+1}.{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i}(pf+(1-p)f^{\prime }\left)\right(x_{1:n+1},x_{n+2:\infty }\left)\right)$

$\ge {lim}_{n\to \infty }p{K}_{:n}\left(x_0\right)(\lambda x_{1:n+1}.{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i}f(x_{1:n+1},x_{n+2:\infty }\left)\right)$

$+(1-p)K_{:n}\left(x_0\right)(\lambda x_{1:n+1}.{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i}{f}^{\prime }(x_{1:n+1},x_{n+2:\infty }\left)\right)$

$=p{lim}_{n\to \infty }{K}_{:n}\left(x_0\right)(\lambda x_{1:n+1}.{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i}f(x_{1:n+1},x_{n+2:\infty }\left)\right)$

$+(1-p){lim}_{n\to \infty }{K}_{:n}\left(x_0\right)(\lambda x_{1:n+1}.{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i}{f}^{\prime }(x_{1:n+1},x_{n+2:\infty }\left)\right)$

$=p{K}_{:\infty }\left(x_0\right)\left(f\right)+(1-p)K_{:\infty }\left(x_0\right)\left(f^{\prime }\right)$

Concavity is shown. Now for 1-Lipschitzness. Fix an arbitrary $n$. Our proof target is:

$|K_{:n}\left(x_0\right)(\lambda x_{1:n+1}.{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i}f(x_{1:n+1},x_{n+2:\infty }\left)\right)$

$-K_{:n}\left(x_0\right)(\lambda x_{1:n+1}.{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i}{f}^{\prime }(x_{1:n+1},x_{n+2:\infty }\left)\right)|\le d(f,f^{\prime })$

Now, if we knew:

$d(\lambda x_{1:n+1}.{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i}f(x_{1:n+1},x_{n+2:\infty }),\lambda x_{1:n+1}.{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i}{f}^{\prime }(x_{1:n+1},x_{n+2:\infty }\left)\right)\le d(f,f^{\prime })$

Then we'd be able to pair that with the 1-Lipschitzness of $K_{:n}\left(x_0\right)$ to get our desired result. So let this be our new proof target. Reexpressing it a bit, it's:

$\forall x_{1:n+1}\in {\prod }_{i=1}^{i=n+1}{X}_i:|{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i}f(x_{1:n+1},x_{n+2:\infty })$

$-{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i}f(x_{1:n+1},x_{n+2:\infty })|\le d(f,f^{\prime })$

Now, if two functions are only $d(f,f^{\prime })$ apart, then their minimal values over a compact set can only be $d(f,f^{\prime })$ apart at most. Let your compact set be ${\prod }_{i=1}^{i=n+1}\left\{x_i\right\}\times {\prod }_{i=n+2}^{\infty }{C}_i$ to see the result. Thus, we have proved our proof target and we have the result that, for all $n,f,f^{\prime }$

$|K_{:n}\left(x_0\right)(\lambda x_{1:n+1}.{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i}f(x_{1:n+1},x_{n+2:\infty }\left)\right)$

$-K_{:n}\left(x_0\right)(\lambda x_{1:n+1}.{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i}{f}^{\prime }(x_{1:n+1},x_{n+2:\infty }\left)\right)|\le d(f,f^{\prime })$

Since they're always only $d(f,f^{\prime })$ apart, the same property transfers to the limit, so we get:

$|K_{:\infty }\left(x_0\right)\left(f\right)-K_{:\infty }\left(x_0\right)\left(f^{\prime }\right)|\le d(f,f^{\prime })$

And so, $K_{:\infty }\left(x_0\right)$ is 1-Lipschitz for all $x_0$. This takes care of the Lipschitzness condition on infradistributions and the uniform Lipschitz bound condition on infrakernels. All that's left is the compact almost-support condition on infradistributions and the compact-shared compact almost-support condition on infrakernels, and the pointwise convergence condition on infrakernels.

Part 7: This will use our result from Part 4 about making a suitable compact set. Our desired thing we want to prove is that

$\forall C^{X_0}\in K\left(X_0\right),ϵ>0\exists C_{ϵ}\subseteq {\prod }_{i=1}^{\infty }{X}_i\forall x_0\in C^{X_0},f,f^{\prime }\in C_B\left({\prod }_{i=1}^{\infty }{X}_i\right):$

$f_{\downarrow C_{ϵ}}=f_{\downarrow C_{ϵ}}^{\prime }\to |K_{:\infty }\left(x_0\right)\left(f\right)-K_{:\infty }\left(x_0\right)\left(f^{\prime }\right)|\le ϵd(f,f^{\prime })$

Because this property is exactly the compact-shared compact-almost-support property for $K_{:\infty }$.

So, here's our plan of attack. Fix our $C^{X_0}$ and $ϵ$, so our target is

$\exists C_{ϵ}\subseteq {\prod }_{i=1}^{\infty }{X}_i\forall x_0\in C^{X_0}f,f^{\prime }\in C_B\left({\prod }_{i=1}^{\infty }{X}_i\right):$

$f_{\downarrow C_{ϵ}}=f_{\downarrow C_{ϵ}}\to |K_{:\infty }\left(x_0\right)\left(f\right)-K_{:\infty }\left(x_0\right)\left(f^{\prime }\right)|\le ϵd(f,f^{\prime })$

Let our $C_{ϵ}$ be $C_{1:\infty }[C^{X_0},ϵ]$, as defined from Part 4, and let $x_0$ be arbitrary in $C^{X_0}$ and $f,f^{\prime }$ be arbitrary and agree with each other on $C_{1:\infty }[C^{X_0},ϵ]$. Then our goal becomes

$|K_{:\infty }\left(x_0\right)\left(f\right)-K_{:\infty }\left(x_0\right)\left(f^{\prime }\right)|\le ϵd(f,f^{\prime })$

Now, letting our chosen defining sequence be $C_i[C^{X_0},ϵ]$, our goal is to show

$\forall n:|K_{:n}\left(x_0\right)(\lambda x_{1:n+1}.{inf}_{x_{n+2:\infty }\in C_{n+2:\infty }[C^{X_0},ϵ]}f(x_{1:n+1},x_{n+2:\infty }\left)\right)$

$-K_{:n}\left(x_0\right)(\lambda x_{1:n+1}.{inf}_{x_{n+2:\infty }\in C_{n+2:\infty }[C^{X_0},ϵ]}{f}^{\prime }(x_{1:n+1},x_{n+2:\infty }\left)\right)|\le ϵd(f,f^{\prime })$

If we could do this, all the approximating points being only $ϵ$ away from each other shows that the same property transfers to the limit and we get our result. Accordingly, let $n$ be arbitrary, and our proof goal is now:

$|K_{:n}\left(x_0\right)(\lambda x_{1:n+1}.{inf}_{x_{n+2:\infty }\in C_{n+2:\infty }[C^{X_0},ϵ]}f(x_{1:n+1},x_{n+2:\infty }\left)\right)$

$-K_{:n}\left(x_0\right)(\lambda x_{1:n+1}.{inf}_{x_{n+2:\infty }\in C_{n+2:\infty }[C^{X_0},ϵ]}{f}^{\prime }(x_{1:n+1},x_{n+2:\infty }\left)\right)|\le ϵd(f,f^{\prime })$

Hypothetically, if we were able to show:

$\forall x_{1:n+1}\in C_{1:n+1}[C^{X_0},ϵ]:{inf}_{x_{n+2:\infty }\in C_{n+2:\infty }[C^{X_0},ϵ]}f(x_{1:n+1},x_{n+2:\infty })$

$={inf}_{x_{n+2:\infty }\in C_{n+2:\infty }[C^{X_0},ϵ]}{f}^{\prime }(x_{1:n+1},x_{n+2:\infty })$

Then we could conclude the two functions were equal on the set $C_{1:n+1}[C^{X_0},ϵ]$, which, from Part 4, is an $ϵ$-almost support for all the $K_{:n}\left(x_0\right)$ where $x_0\in C^{X_0}$, and it would yield our result. So our proof target is now

$\forall x_{1:n+1}\in C_{1:n+1}[C^{X_0},ϵ]:{inf}_{x_{n+2:\infty }\in C_{n+2:\infty }[C^{X_0},ϵ]}f(x_{1:n+1},x_{n+2:\infty })$

$={inf}_{x_{n+2:\infty }\in C_{n+2:\infty }[C^{X_0},ϵ]}{f}^{\prime }(x_{1:n+1},x_{n+2:\infty })$

Accordingly, let $x_{1:n+1}$ be arbitrary within said set, turning our proof target into:

${inf}_{x_{n+2:\infty }\in C_{n+2:\infty }[C^{X_0},ϵ]}f(x_{1:n+1},x_{n+2:\infty })={inf}_{x_{n+2:\infty }\in C_{n+2:\infty }[C^{X_0},ϵ]}{f}^{\prime }(x_{1:n+1},x_{n+2:\infty })$

However, $f$ and $f^{\prime }$ are equal on the set $C_{1:\infty }[C^{X_0},ϵ]$, which breaks down as $C_{1:n+1}[C^{X_0},ϵ]\times C_{n+2:\infty }[C^{X_0},ϵ]$. And we know that $x_{1:n+1}\in C_{1:n+1}[C^{X_0},ϵ]$, and since the second part is being selected from $C_{n+2:\infty }[C^{X_0},ϵ]$, the two values are equal, and we've proved our proof target. Thus, we have compactly-shared  compact almost-support, and that's the second condition for $K_{:\infty }$ to be an infrakernel, as well as the last condition needed for all the $K_{:\infty }\left(x_0\right)$ to be infradistributions. Just pointwise convergence left!

Part 8: Alright, for showing pointwise convergence, we'll need a lemma. We need to show:

$\forall C^{X_0}\subseteq X_0,\overline{C}\in {\prod }_{i=1}^{\infty }K\left(X_i\right),ϵ>0,f\in C_B\left({\prod }_{i=1}^{\infty }{X}_i\right)\exists n\in N\forall m\in N,x_0\in C^{X_0}:$

$|K_{:n}\left(x_0\right)(\lambda x_{1:n+1}.{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i}f(x_{1:n+1},x_{n+2:\infty }\left)\right)$

$-K_{:n+m}\left(x_0\right)(\lambda x_{1:n+m+1}.{inf}_{x_{n+m+2:\infty }\in {\prod }_{i=n+m+2}^{\infty }{C}_i}f(x_{1:n+m+1},x_{n+m+2:\infty }\left)\right)|\le ϵ(2||f||+1)$

This is roughly saying that the rate of convergence of $K_{:n}\left(x_0\right)\left(f\right)$ to $K_{:\infty }\left(x_0\right)\left(f\right)$ is uniform over compact sets. Let's begin. Let $C^{X_0},\overline{C},ϵ,f$ be arbitrary, so our proof target is now:

$\exists n\in N\forall m\in N,x_0\in C^{X_0}:|K_{:n}\left(x_0\right)(\lambda x_{1:n+1}.{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i}f(x_{1:n+1},x_{n+2:\infty }\left)\right)$

$-K_{:n+m}\left(x_0\right)(\lambda x_{1:n+m+1}.{inf}_{x_{n+m+2:\infty }\in {\prod }_{i=n+m+2}^{\infty }{C}_i}f(x_{1:n+m+1},x_{n+m+2:\infty }\left)\right)|\le ϵ(2||f||+1)$

Here's what we'll do. Because we have a $C^{X_0}$ and $ϵ$, we can craft our sequence $C_i[C^{X_0},ϵ]$ of compact $ϵ$-supports. Now, on the set:

${\prod }_{i=1}^{\infty }\left(C_i\right[C^{X_0},ϵ]\cup C_i)$

which is compact, we know that $f$ is uniformly continuous, so there's some $\delta$ you need to go to to ensure that function values are only $ϵ$ away from each other, which translates into an $n$ via $n:={\log }_2\left(\delta \right)$. Any two inputs which only differ beyond that point will be only $\delta$ apart and can only get an $ϵ$-difference in values from $f$. Now that that's defined, let $m$ and $x_0\in C^{X_0}$ be arbitrary. Our proof target is now to show:

$|K_{:n}\left(x_0\right)(\lambda x_{1:n+1}.{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i}f(x_{1:n+1},x_{n+2:\infty }\left)\right)$

$-K_{:n+m}\left(x_0\right)(\lambda x_{1:n+m+1}.{inf}_{x_{n+m+2:\infty }\in {\prod }_{i=n+m+2}^{\infty }{C}_i}f(x_{1:n+m+1},x_{n+m+2:\infty }\left)\right)|\le ϵ(2||f||+1)$

We know from Part 3 that

$K_{:n}\left(x_0\right)(\lambda x_{1:n+1}.{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i}f(x_{1:n+1},x_{n+2:\infty }\left)\right)$

$=K_{:n+m}\left(x_0\right)(\lambda x_{1:n+m+1}.{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i}f(x_{1:n+1},x_{n+2:\infty }\left)\right)$

So let's make that substitution, turning our proof target into

$|K_{:n+m}\left(x_0\right)(\lambda x_{1:n+m+1}.{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i}f(x_{1:n+1},x_{n+2:\infty }\left)\right)$

$-K_{:n+m}\left(x_0\right)(\lambda x_{1:n+m+1}.{inf}_{x_{n+m+2:\infty }\in {\prod }_{i=n+m+2}^{\infty }{C}_i}f(x_{1:n+m+1},x_{n+m+2:\infty }\left)\right)|\le ϵ(2||f||+1)$

Assuming hypothetically we were able to show that:

$\forall x_{1:n+m+1}\in C_{1:n+m+1}[C^{X_0},ϵ]:|{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i}f(x_{1:n+1},x_{n+2:\infty })$

$-{inf}_{x_{n+m+2:\infty }\in {\prod }_{i=n+m+2}^{\infty }{C}_i}f(x_{1:n+m+1},x_{n+m+2:\infty })|\le ϵ$

Then, because $C_{1:n+m+1}[C^{X_0},ϵ]$ is an $ϵ$-almost-support for $K_{:n+m}\left(x_0\right)$ as long as $x_0\in C^{X_0}$ (which we know to be the case), we could apply Lemma 2 to get our desired proof target.

The "Lipschitz constant times deviation over compact set" part would make one $ϵ$, and the "degree of almost-support times difference in the two functions" part would make $2ϵ||f||$ in the worst-case due to being an $ϵ$-almost-support and the worst possible case where the two functions differ by $2||f||$ somewhere.

So now our proof target is:

$\forall x_{1:n+m+1}\in C_{1:n+m+1}[C^{X_0},ϵ]:|{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i}f(x_{1:n+1},x_{n+2:\infty })$

$-{inf}_{x_{n+m+2:\infty }\in {\prod }_{i=n+m+2}^{\infty }{C}_i}f(x_{1:n+m+1},x_{n+m+2:\infty })|\le ϵ$

Thus, let $x_{1:n+m+1}$ be arbitrary within the appropriate set, so our proof target is now:

$|{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i}f(x_{1:n+1},x_{n+2:\infty })$

$-{inf}_{x_{n+m+2:\infty }\in {\prod }_{i=n+m+2}^{\infty }{C}_i}f(x_{1:n+m+1},x_{n+m+2:\infty })|\le ϵ$

Because we're minimizing over a compact set in both instances, we can consider the infs to be attained by a $x_{n+2:\infty }$ and a $x_{n+m+2:\infty }$, so our proof target is now

$|f(x_{1:n+1},x_{n+2:\infty })-f(x_{1:n+m+1},x_{n+m+2:\infty })|\le ϵ$

Now, because $x_{1:n+1}\in C_{1:n+1}[C^{X_0},ϵ]$, and $x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i$, we have:

$(x_{1:n+1},x_{n+2:\infty })\in C_{1:n+1}[C^{X_0},ϵ]\times {\prod }_{i=n+2}^{\infty }{C}_i$

$={\prod }_{i=1}^{i=n+1}{C}_i[C^{X_0},ϵ]\times {\prod }_{i=n+2}^{\infty }{C}_i\subseteq {\prod }_{i=1}^{\infty }\left(C_i\right[C^{X_0},ϵ]\cup C_i)$

And also, because $x_{1:n+m+1}\in C_{1:n+m+1}[C^{X_0},ϵ]$, and $x_{n+m+2:\infty }\in {\prod }_{i=n+m+2}^{\infty }{C}_i$,

$(x_{1:n+m+1},x_{n+m+2:\infty })\in C_{1:n+m+1}[C^{X_0},ϵ]\times {\prod }_{i=n+m+2}^{\infty }{C}_i$

$={\prod }_{i=1}^{i=n+m+1}{C}_i[C^{X_0},ϵ]\times {\prod }_{i=n+m+2}^{\infty }{C}_i\subseteq {\prod }_{i=1}^{\infty }\left(C_i\right[C^{X_0},ϵ]\cup C_i)$

So, both inputs lie in the relevant compact set, and they agree on $x_{1:n+1}$, so the inputs are only $\delta$ apart, so they only have value differing by $ϵ$, and we're done, our desired result of 

$\forall C^{X_0}\subseteq X_0,\overline{C}\in {\prod }_{i=1}^{\infty }K\left(X_i\right),ϵ>0,f\in C_B\left({\prod }_{i=1}^{\infty }{X}_i\right)\exists n\in N\forall m\in N,x_0\in C^{X_0}:$