The following Theorem I have thought for a long time. I feel like there seems to be no simple solution for it. Does anyone have any ideas or hints for this?
Let $H$ be a previsible process. $X$ be a local martingale. Suppose $H$ is integrable with respect to $X$. If $H\cdot X$ is bounded below by a constant, then $H\cdot X$ is a local martingale.
If $H$ is locally bounded, then $H\cdot X$ is certainly a local martingale. So the theorem basically deals with the case $H$ is not locally bounded. It is pretty significant, since admissible financing strategies requires the payoff to be bounded below. With this it basically saying the payoff $H\cdot X$ is a local martingale.
EDIT:
Definition of integrability for local martingale is
$H$ integrable w.r.t $X$ (when $X$ is a local martingale) means there exists a stopping time sequence $T_n\rightarrow\infty$, $\mathbb{E}\left(\int H^2 \ d[X^{T_n-}, X^{T_n-}]\right)<\infty$. (This is the definition from Protter's Stochastic differential equation textbook, integrable in $T_n-$ sense).
Proof
One possible proof I can think of(after extracting several proof methods from several theorems):
First, $H\cdot X$ is a $\sigma$-martingale since the integrator $X$ is a local martingale. WLOG, assume $H\geq 0$.
Let $$A=\sum_{s\leq t}1_{(|\Delta X|>1 \ \text{or} \ |\Delta (H\cdot X)|>1)}\Delta X$$ Since $X, H\cdot X$ are cadlag (which are regular), $A$ is of finite variation. And further we by stopping with $Var(A)\geq n$, $A$ is locally integrable. Take $N_2 = A-A^p$ ($A^p$ is the compensator, so $N_2$ is a local martingale). Take $N_1 = X-N_2$. We show $H\cdot N_1, H\cdot N_2$ are both local martingales.
Verify integrability: By the definition of $A$, the Stieljies integral $\int H\ dA$ exists. Since $H$ is non-negative, by compensator's property and suitable stopping $T_n-$, $H$ is integrable w.r.t. $A^p$ ($\mathbb{E}(H\cdot A) = \mathbb{E}(H\cdot A^p)$. By stopping $T_n-$ we can make these expectation finite). Vector space structure ensures $H$ integrable w.r.t. to $N_2$ and hence $N_1$ (in $T_n-$ sense).
We show $H\cdot N_1$ is a local martingale. Property of compensator: $^p(\Delta A) = \Delta (A^p)$ ($^p X$ is the predictable prejection of $X$). Define \begin{align*} Y = 1_{(|\Delta X|\leq 1 \ \text{and} \ |\Delta (H\cdot X)|\leq 1)}\Delta X \end{align*} Then $^p (\Delta A) = - ^p Y$ (use the fact, when $X$ is a local martingale, $^p (\Delta X) = 0$.) \begin{align*} \Delta N_1 &= \Delta X - \Delta A + \Delta (A^p)\\ &= Y - ^p Y \end{align*} since $|Y|\leq 1$, $|\Delta N_1|\leq 2$ (predictive projection can be written as conditional expectation of the original). This shows that $N_1$ has bounded jumps. Similar argument shows $|H\Delta N_1|\leq 2$. By suitable stopping, we can assume $N_1$ is square integrable martingale with limit in $L^2$. Define $H_n$ as $H 1_{|H|\leq n}$. Because $|H\Delta X|\leq 2$, the integrabilty of $H$ can be extended from $T-$ to $T$ for any stopping time $T$. Thus, we can assume $H$ is integrable w.r.t. $N_1$ in $\mathcal{H}^2$ sense (i.e. not in stopping $T_n-$ sense).
By the definition of Stochastic integral, $H\cdot N_1$ is the limit of $H_n\cdot N_1$ in $\mathcal{H}^2$, this shows $H\cdot N_1$ is a square integrable martingale.
Next, we show $H\cdot N_2$ is a local martingale. Recall the notation $H_n=H 1_{(|H|\leq n)}$. $H_n\cdot N_2$ are local martingales. By stopping $T_n = \inf\{t, H\cdot X\geq n\}$, and $H\cdot X$ is bounded below, \begin{align*} -c \leq \Delta (H\cdot X) = H \Delta X \end{align*} where $-c$ is the lower bounds for $H\cdot X$. By the definition of $N_1,N_2$, we can get $H\Delta N_2\geq -c-2$. This shows $(H\Delta N_2)^-\leq c+2$ where $(\cdot )^-$ denote the negative parts of a function. Clearly, $(H_n\Delta N_2)^-\leq c+2$.
By stopping, we assume they are $N_2$ are integrable process (i.e. $\mathbb{E}(\int_0^\infty d|N_2|)<\infty$). $H_n\cdot N_2$ are local martingale with \begin{align*} \mathbb{E}\sup|H_n\cdot N_2|\leq n\mathbb{E}\left(\int_0^\infty d|N_2|\right)<\infty \end{align*} This shows $H_n\cdot N_2$ are true martingales.
This implies $\mathbb{E}(H_n\cdot N_2)^+ = \mathbb{E}(H_n\cdot N_2)^-$. By stopping on $T = \inf\{t, \int_0^t |H_t| d|N_2|\geq c\}$, we can bound (LHS is a Stieljies integral) \begin{align*} (H_n\cdot N_2)_T^- \leq \int_0^{T-}|H_t|d|N_2| + (H_n\Delta X)^- \leq 2c+2 \end{align*} Consequently \begin{align*} \mathbb{E}|H_n\cdot N_2|_T<4c+4 \end{align*} Fatou's Lemma shows \begin{align*} \mathbb{E}|H\cdot N_2|_T<4c+4 \end{align*} Next, \begin{align*} |H_n\cdot N_2|_T &\leq |H_n\cdot N_2|_{T-}+|H_n\Delta N_2|_T \\ &\leq 2\int_0^{T-}|H|d|N_2| + |H\cdot N_2|_T \end{align*} The RHS has finite expectation.
By martingale property of $H_n\cdot N_2$ \begin{align*} \mathbb{E}[(H_n\cdot N_2)_T|\mathcal{F}_t] = (H_n\cdot N_2)_{T\land t} \end{align*} Apply $L_1$ convergence on both sides obtain the result.
