Theorem 4.2.8 in Durrett: Probability Theory and Examples states
Let $X_n, n \geq 0$ be a supermartingale. If $H_n \geq 0$, is predictable and each $H_n$ is bounded then $(H \cdot X)_n := \sum_{m=1}^{n}H_m(X_m-X_{m-1})$ is a supermartingale.
While I know an example showing that this statement does not hold when $H_n$ is not bounded, I cannot see where we use boundedness in the proof. The proof is based on this equation: $$\mathbb{E}[(H\cdot X)_{n+1} | \mathcal{F}_n ] = (H\cdot X)_{n} + \mathbb{E}[H_{n+1}(X_{n+1}-X_{n}) | \mathcal{F}_n ] = (H\cdot X)_{n} + H_{n+1}\mathbb{E}[(X_{n+1}-X_{n}) | \mathcal{F}_n ] \leq (H\cdot X)_{n} $$ since $H_{n+1} \geq 0$ and $\mathbb{E}[(X_{n+1}-X_{n}) | \mathcal{F}_n ] \leq 0$.
Now the properties of conditional expectation we use in this equation (linearity and that we can factor $H_{n+1}$ out because it is $\mathcal{F}_n$ measurable) do require $(H\cdot X)_{n+1}$ and $H_{n+1}$ to be integrable. Is boundedness just a way to guarantee that these two conditions hold, or is boundedness actually necessary?
I believe that boundedness must appear somewhere else since the gambling system where start with one dollar, we double the stakes whenever we loose and stop playing once we have won our dollar back, has each $H_n$ and $(H \cdot X)_{n}$ integrable, even though the $H_n$ are not bounded.
The example I mean comes from Durrett (slightly modified, see comment at the end):
Let $X_n = \sum_{i=1}^{n} \chi_i$ where $\chi_i = 1$ with probability $p$ and $-1$ with probability $1-p$. Let $H_n = 2H_{n-1}$ if $X_{n-1} = - 1$, $H_n = 0$ if $X_{n-1} = 1$. Then $\mathbb{P}(H_n = 2^k) = p^k$ so $H_n$ is unbounded, but for each $n$, $H_n$ and $(H \cdot X)_{n}$ are integrable. Now since $\mathbb{P}(H_n < \infty)=1$, $\mathbb{E}[ \lim_{n\to \infty} (H \cdot X)_{n}]=1$, i.e. we are sure to win our dollar back. So eventhough we started with a strict supermartingale (if $p<\frac{1}{2}$) we ended up not loosing money, i.e. we kind of cheated the system. (I modified the example a little by setting $H_n=0$ after we win so that we do end up really winning the dollar back. In Durrett's example we start again betting 1 dollar once we have a net gain of 1, i.e. $H_n = 2H_{n-1}$ if $X_{n-1} = - 1$, $H_n = 1$ if $X_{n-1} = 1$.)
I now realize that this does not contradict the theorem. I was led to think that because after stating this example Durrett says "This system seems to provide us with a "sure thing" as long as $\mathbb{P}(\chi_m=1)>0$. However the next results says there is no system for beating an unfavorable game." So my new question is: can a gambling system prevent us from loosing while playing a strict supermartingale?
Since $X_n$ is an (integrable) supermartingale, boundedness of $H_n$ guarantees that $H_nX_n$ is integrable.
Note that $x^{-1/2}$ is integrable on $[0,1]$ with respect to Lebesgue measure , but $x^{-1/2} \cdot x^{-1/2}$ is not.