From information theory, suppose that there exists a bit stream $(b_i)$ of nonstandard length $LK$ such that for any nonstandard sample of $M\ll L$ consecutive substrings of length $K$ the sequence of $MK$ bits has Kolmogorov complexity $MK(1+\epsilon)$ bits in some simple restricted LISP like language, where $\epsilon \cong 0$. From the theory, this is near maximal complexity since every finite string has a complexity no more than some constant larger than that finite string, and our strings are finite but nonstandard.

Intuitively, we have a bit sequence which is very very long, and when we take very long samples of long bit strings, we can't compress those samples.

Such a bit stream is non-computable, and nonstandard. But does it exist? I don't know. By transfer, it would have to exist if for all standard values of $L,K$ there is a sequence with prefix bitstring of length $MK$  with complexity $MK+C$ for some constant $C$. So suppose it does exist. We can make a random number generator from this sequence easily, in a manner similar to what is done in a computer.

The function $\xi^*(n)$ takes $K=log_2(N)\cong \infty$ sequential bits starting at location $nK$ and forms the number $(b)/2^K \in [0,1)$ in other words interpret the bits $(b)$ as an integer and divide by $2^K=N$. This number will be a nonstandard rational number whose binary expansion has $K$ bits (a nonstandard number of bits). The standard part of this number will in general be irrational.  The values of $\xi^*$ all lie on a lattice whose lattice spacing is $1/N$. The function $\xi^*(n)$ is clearly a nonstandard function, but we can make it output standard values by taking the standard part of its output, then it will be a function $\xi(n) = st(\xi^*(n)) : \mathbb N \rightarrow [0,1]$.

Now, we'd like to show that the nonstandard function $\xi^*$ has the property of being "near-uniformly-distributed" on all micro-intervals $[b,a] \subset [0,1]$ of sufficiently long but infinitesimal length. So we need a reasonable definition of "near-uniformly-distributed". Suppose that we take a nonstandard sample of $M$ successive outputs of $\xi^*$ and we count the relative frequency that the output is in any infinitesimal interval $[a,b]\subset [0,1]$. We require that this frequency be $(b-a)(1+\eta)$ with $\eta \cong 0$. In other words, on any interval which includes a standard number of infinitesimal lattice points, we need to have near uniformity for large enough sample size $M$.

Suppose that $\eta$ is positive and appreciable for some micro- interval. Then the output of the bit sequence produces numbers in that interval significantly more often than in other intervals. A compressor program for the original bitstring would be able to take advantage of this fact. In fact we could output the bitstring in chunks of length $K$ followed by a 0 if we are going to output another chunk, or a 1 followed by the address within the micro-interval of the lattice point we're going to output. These addresses need to be of length $K+\log_2(b-a) = K(1-\epsilon_K)$ so we save a lot of bits by knowing the micro-region we're going to output.

For $M$ chunks of length $K$ we could re-encode the stream as approximately $M(1-(b-a)(1+\eta))(K+1) + M(b-a)(1+\eta)K(1-\epsilon_K)$ bits using this method. As a fraction of $MK$ bits that's: $1-(b-a)(1+\eta) \epsilon_K$. But $(b-a)$ is infinitesimal and so we haven't saved much unless $\eta$ is nonstandard. This shows that if we have a point mass component to the distribution that we can compress the original sequence. So we've shown that $\xi(n)$ has a standard density on $[0,1]$.

It'd be nice to take it further and show that this standard density is very close to uniform, perhaps I can do that or perhaps not. I'll make an effort next time.