Why do we require a stochastic integral to be a martingale? Kuo, in his book "Introduction to Stochastic Integration" ( pages 37-41 ) to justify certain choices in the definition of a stochastic integral:
$$I(t)=\int_0^t f(s,\omega)dB(s,\omega)$$
where $B$ is a Brownian motion and $f$ a stochastic process (with certain characteristics), emphasizes that some decisions are done so that $I(t)$ is a martingale.
For example:

*

*Kuo observes that:

$$\int_0^t B(1,\omega)dB(s,\omega)$$
should not be defined by any integration scheme, since the value that we expect from a reasonable integration theory would be:
$$I(t,\omega)=B(1,\omega)B(t,\omega)$$
, and this is not a martingale, since the expectation value depends on $t$. From here Kuo "decides" to restrict $f$ to be adapted ;

*

*An other example is, if we try to calculate $\int_0^t B(t)dB(t)$ with a discrete approximation:

$$\int_0^t B(t)dB(t) \sim I^{\Delta}=\sum_i B(\tau_i)(B(t_{i})-B(t_{i-1}))$$
Only when the evaluation point $\tau_i$ is equal to $t_{i-1}$, than the limit of $I^{\Delta}$ in $L^2$ defines a martingale, and from here Kuo decides that the evaluation point should be on the left ( so the Ito and not Stratonovich choice ).
I understand from the first chapters that the Wiener or Stjeltes integral ( for $f$ deterministic in $L^2$ or of bounded variation respectively ) return an integral w.r.t. $B$ that is a martingale. So one could proceeed "by analogy". But I do not think this is the real reason. So here the questions:

*

*Why do we require (oder at least Kuo) $I(t)$ to be a martingale ?

*What are some applications of the martingale property (or useful properties of the resulting integration scheme resulting from it) in this context ?

 A: I could vaguely argue, from the point of view of real-life experiments, that the martingale property is useful to describe any phenomenon in which "any measurement I make at time $t$ is a 'good' approximation of a measurement I make at time $t+\tau$ for some small $\tau$".
The property $E[X_{t+\tau}\mid \mathcal{F}_t]=X_t$ means that $X_{t+\tau}$ will have same averages or expectations as $X_t$, given the accumulated information in $\mathcal{F}_t$. Implied by this is that  $E[X_t]$ does not change for all $t$. Could you think about some real life phenomenon modeled by a martingale? Perhaps a deteriorating sensor where $X_t$ denotes the error in the measurement at each $t$ for example? Or maybe $X_t$ describes the error a partially intoxicated person makes when asked to walk in a straight line? Even simpler, consider a discrete process $\{X_n\}$ where $X_n$ denotes your gain after tossing a fair coin $n$ times, assuming you get 1 if heads and -1 if tails. This process can be shown to be a Martingale, which makes intuitive sense.
In your question, you also ask why we use a property such as $\{f_t\}$ is $\mathcal{F}_t$ adapted. Note that $X_t$ being $\mathcal{F}_t$-adapted implies that $E[X_t\mid \mathcal F_t]=X_t$, ie the 'information accumulated up to time $t$ in $\mathcal{F}_t$ is sufficient to determine the value of $f_t$ on all $A\in\mathcal{F}_t$'. So for example if $\mathcal{F}_t=\bigcup_{0\leq s\leq t}\sigma(B_s)$, where $\{B_t\}_{t\geq 0}$ is Brownian motion, then you should be able to determine the value of $f_t=f(B_t)$ for some Borel-measurable function $f$ (for example $ f_t=B_t^2$ or $f_t=e^{B_t}$) form the information in $\mathcal{F}_t$, because $\mathcal{F}_t$ contains $\sigma(B_t)$, which determines $B_t$ almost everywhere.
So even before mentioning the integral, the martingale property is quite useful. Recall also that when choosing the left endpoint in the Ito integral definition of a simple process $\Delta(t)$
$$\int_0^t\Delta(s)dB_s = \sum_{j=0}^{k-1}\Delta(t_j)(B_{t_{j+1}}-B_{t_j})+\Delta(t)(B_{t}-B_{t_k}),\label{1}\tag{1}$$
you do not 'peak into the future', everything is determined from $\mathcal{F}_t$. There is nice point of view in Stochastic Calculus for Finance II: Continuous-Time Models by Shreve that one can adopt to understand the Ito integral of simple processes:

Regard $B_t$ as the
price per share of an asset at time $t$. (Since Brownian motion can take negative
as well as positive values, it is not a good model of the price of a limited­
liability asset such as a stock. For the sake of this illustration, we ignore that
issue.) Think of $t_0,t_1,\ldots,t_{n-1}$ as the trading dates in the asset, and think
of $\Delta(t_0) , \Delta(t_1 ), \ldots , \Delta (t_n - 1 )$ as the position (number of shares) taken in the
asset at each trading date and held to the next trading date ($\Delta$ is a simple process). The Ito integral in \eqref{1} is defined as the gain from trading in the martingale $B_t$. A martingale has no tendency to rise or fall, and hence it is to be
expected that $I(t)$, thought of as a process in its upper limit of integration $t$,
also has no tendency to rise or fall.

You can follow up form here.
