Measurability of stopping times

Let $(\mathcal{F}_t)_{t \geq 0}$ be a filtration and $\tau$ an $\mathcal{F}_t$-stopping time, i.e. $\{\tau \leq t\} \in \mathcal{F}_t$ for all $t \geq 0.$

For a fixed constant $c >0$ the mapping $\varrho(\omega) := c$ defines an $\mathcal{F}_t$-stopping time. Then $\tau-\varrho = \tau-c$ is, in general, not a $\mathcal{F}_t$-stopping time since

$$\{\tau-\varrho \leq t\} = \{\tau \leq t+c\}$$

is in $\mathcal{F}_{t+c}$ but not necessarily in $\mathcal{F}_t$. A similar thing happens if we consider $\varrho \cdot \tau$ and $\varrho:=c >1$.

Intuitively, one should think about a stopping time in conjunction with a process. It's nice to think about a process as a stock price, and a stopping time as a strategy for deciding when to sell.

For instance, imagine a stock whose price at time $t=0$ is 100. let $\tau$ be the first time the stock price hits 200 (i.e. $\tau = \inf\{t : X_t = 200\}$). Similarly $\rho$ be the first time the price reaches 50. So $\tau$ itself is a reasonable strategy: "sell when the price reaches or exceeds 200". You could actually do that, so $\tau$ is a stopping time. $\tau+5$ is also a reasonable strategy: "Wait until the price hits 200, wait 5 more days, then sell." So $\tau+5$ is also a stopping time. $\tau-5$ is not a stopping time: "Sell five days before the stock hits 200." If the stock hits 200 on day 8, you won't know that until day 8, at which point you'll see you should have sold on day 3, but by then it's too late.

So $\tau+\rho$ is a stopping time: "Wait until the stock has hit both prices, add the times, and sell at that time." If it hits 200 on day 8 and 50 on day 9, you are supposed to wait until day 17 to sell. You could actually do that. It's a stopping time, though a stupid one because there is no particular meaning to adding absolute times to get an absolute time.

$\tau - \rho$ is not a stopping time for reasons similar to $\tau-5$ (though it is still stupid because subtracting absolute times isn't meaningful as an absolute time). For $\tau\rho$, consider what happens if $\tau = 1/2$ and $\rho = 3/4$. (And it's even stupider because multiplying two times doesn't give a time at all; the units are wrong.)