Prove that an Itô integral is a martingale

Background Q. Prove that an Itô integral with respect to Brownian motion is a martingale My Attempt To prove that an Itô integral with respect to Brownian motion is a martingale (representing a "fair game" with no predictable drift), we look at the properties of the Itô integral stochastic process: $I_t = \int_0^t H_s \, dW_s$ where $W_s$ is a standard Brownian motion and $H_s$ is a suitably integrable, adapted process (meaning the trader cannot look into the future to pick their strategy $H_s$ ). For $I_t$ to be a martingale, it must satisfy two main conditions for any future time $t$ and present time $s$ (where $s \le t$ ): Integrability: $\mathbb{E}[\vert{}I_t\vert{}] < \infty$ The Martingale Condition: the conditional expectation of the future value given all information up to the present time $s$ must equal the present value: $\mathbb{E}[I_t \mid \mathcal{F}_s] = I_s$ Proof. Let's split the integral into past and future segments. We can split the integral from $0$ to $t$ into two pieces: $\color{blue}{\text{from}\ 0 \ \text{to}\ s}$ , and $\color{red}{\text{from}\ s \ \text{to}\ t}$ : $I_t = \int_0^t H_u \, dW_u = \color{blue}{\int_0^s H_u \, dW_u} + \color{red}{\int_s^t H_u \, dW_u}$ Notice that the first part ( $\int_0^s H_u \, dW_u$ ) is simply $I_s$ , and it is already fully "known" (measurable) at time $s$ . Take the conditional expectation at time $s$ : $\mathbb{E}[I_t \mid \mathcal{F}_s] = \mathbb{E}\left[ \color{blue}{\int_0^s H_u \, dW_u} + \color{red}{\int_s^t H_u \, dW_u} \Bigg\vert{} \mathcal{F}_s \right]$ Split by linearity: $\mathbb{E}[I_t \mid \mathcal{F}_s] = \mathbb{E}\left[ \color{blue}{\int_0^s H_u \, dW_u} \Bigg\vert{} \mathcal{F}_s \right] + \mathbb{E}\left[ \color{red}{\int_s^t H_u \, dW_u} \Bigg\vert{} \mathcal{F}_s \right]$ Evaluate both parts: the past: since $\int_0^s H_u \, dW_u$ ( $I_s$ ) has already happened by time $s$ , its conditional expectation given current information is just itself: $\mathbb{E}\left[ \color{blue}{\int_0^s H_u \, dW_u} \Bigg\vert{} \mathcal{F}_s \right] = \color{blue}{I_s}$ the future: by the fundamental property of Itô integrals over future intervals ( $s$ to $t$ ), the expected value of future Brownian increments given current information is $0$ (the definition of a fair game with no drift): $\mathbb{E}\left[ \color{red}{\int_s^t H_u \, dW_u} \Bigg\vert{} \mathcal{F}_s \right] = \color{red}{0}$ Combine both of them: $\mathbb{E}[I_t \mid \mathcal{F}_s] = \color{blue}{I_s} + \color{red}{0} = I_s$ Because the conditional expectation of the future Itô integral equals its current value ( $\mathbb{E}[I_t \mid \mathcal{F}_s] = I_s$ ), the Itô integral is a martingale. This mathematically guarantees that we cannot systematically make or lose expected profit over time using a non-anticipating trading strategy in a frictionless market. $\blacksquare$ My Question If $\mathbb{E}[I_t \mid \mathcal{F}_s] = I_s$ means that the expected future value equals our current value, not zero. Therefore, if our current Itô integral value is $I_s = 0$ , the expected future value is $0$ ? if our current value is positive (e.g., I_s = \\\500$in profit) or negative (e.g.,$I_s = -\$200$in loss), our expected future value is whatever our current value is ($500$or$-200$)? it's a "fair game" because our expected change from right now into the future is zero ($\mathbb{E}[I_t - I_s \mid \mathcal{F}_s] = 0$ ). Therefore, we don't have an automatic upward drift or downward pull, as we are expected to stay right where we currently are on average? Did I get this correct?