Suppose you replace the usual success conditions for a supermartingale (lim sup is infinite) with the requirement that the actual limit is infinite, e.g. a supermartingale $B$ succeeds on $X \in 2^\omega$ just if


$$\lim\_{n \to \infty} B(X\restriction\_n) = \infty$$


I'm 90% sure that you don't get a 'valid' randomness notion because any sufficiently generic real should qualify. However, if I had to guess I'd think this notion turns out to be equivalent to Kurtz 'randomness' (avoids all $\Pi^0\_1$ null sets). Since I'm guess this is a known result and my brain is feeling super foggy I thought I would ask



Kurtz randomness and supermartingales with infinite *limit*
Assuming you mean a c.e. (super)martingale, this is just Martin-Löf randomness. What needs to be shown is that, for every non-random real, there is a martingale which limit-succeeds on the real. You can see this from the proof of the equivalence of the martingale definition and the Kolmogorov complexity definition:


Let $U$ be a universal prefix-free machine. For $\sigma \in 2^{<\omega}$, define $M\_\sigma$ to be the (computable) martingale that begins with capital 1 and bets it all on $\sigma$, and bets evenly afterwards. So $M(\tau) = 2^{|\sigma|}$ for any $\tau$ extending $\sigma$. Define $M = \sum\_{U(\rho)=\sigma} 2^{-|\rho|}M\_\sigma$. This is a c.e. martingale (with starting capital $\Omega$). For any non-random $X$, for any $d$, there is an $n$ with $K(X\upharpoonright n) < n - d$. So there is some $\rho$ with $|\rho| < n - d$ and $U(\rho) = X\upharpoonright n$. So $2^{-|\rho|} M\_{X\upharpoonright n}$ is a summand in the definition of $M$, and $2^{-|\rho|}M\_{X\upharpoonright n}(\tau) > 2^d$ for any $\tau$ extending $X\upharpoonright n$. So $\lim\_{n \to \infty} M(X\upharpoonright n) = \infty$.


Note that this construction shows why your generic intuition is wrong: for every $d$, there is a neighborhood of $X$ on which the martingale never again dips below $2^d$. So generics can't get out of it.


