Confusion about a conditional expectation Suppose $X$ is an integrable, positive random variable and define the filtration $(\mathcal{F}_t)_{t\ge 0}$ with $\mathcal{F}_t=\sigma(\{X\le s\} :s\le t)$. I want to calculate $\mathbb{E}[X|\mathcal{F}_t]$, but I seem to get two very different results depending on how I calculate it.
On one hand, I would write $X=\mathbf{1}_{\{X>t\}}X+\mathbf{1}_{\{X \le t\}}X$ and say that both of these terms are $\mathcal{F}_t$-measurable. But then $X$ itself is $\mathcal{F}_t$-measurable, which I think is really strange since that would mean $\sigma(X) \subseteq \mathcal{F}_t $ for every $t \ge 0$ which can't be true.
On the other hand, I have shown that for example
$$\mathbf{1}_{\{X >t\}}\mathbb{E}[X|\mathcal{F}_t] =\mathbb{E}[\mathbf{1}_{\{X >t\}}X|\mathcal{F}_t] = \mathbf{1}_{\{X>t\}}\frac{\mathbb{E}[\mathbf{1}_{\{X>t\}}X]}{\mathbb{E}[\mathbf{1}_{\{X>t\}}]}$$
And similariy on the complement.
I showed this by showing that the right hand side integrates like $\mathbf{1}_{\{X>t\}}X$ on every set from $\{\{X\le s\}:s\le t\}\cup \{\Omega\}$ which is stable under intersections and contains the underlying set $\Omega$, so this should suffice. But both results seem weird to me since in "Introduction to the theory of point processses" by Dayley and Vere-Jones, they claim that
$$\mathbb{E}[X|\mathcal{F}_t]=\mathbf{1}_{\{X\le t\}}X + \mathbf{1}_{\{X>t\}}\frac{\mathbb{E}[\mathbf{1}_{\{X>t\}}X]}{\mathbb{E}[\mathbf{1}_{\{X>t\}}]}$$
Clearly something has gone wrong somewhere, but I cannot see where - which, if any, of the above results are correct?
 A: You said '$X=\mathbf{1}_{\{X>t\}}X+\mathbf{1}_{\{X \le t\}}X$ and say that both of these terms are $\mathcal{F}_t$-measurable': This is not correct. The first term is not $\mathcal{F}_t$-measurable.  If $X(\omega) >1$ for all $\omega$ then $\mathcal F_1=\{\emptyset, \Omega\}$ and $X$ is not meaurable w.r.t. $\mathcal F_1$ unless $X$ is a  constant.
There is no contradiction between what you have obtained and what the book says. Note that $X1_{X\leq t}$ is $\mathcal F_t$ measurable so $E[X1_{X\leq t}|\mathcal F_t]=X1_{X\leq t}$. So your result is compatible with what they have written.
A: The solution
$$\mathbb{E}[X|\mathcal{F}_t]=\mathbf{1}_{\{X\le t\}}X + \mathbf{1}_{\{X>t\}}\frac{\mathbb{E}[\mathbf{1}_{\{X>t\}}X]}{\mathbb{E}[\mathbf{1}_{\{X>t\}}]}$$
is correct.
This means:
$$\mathbb{E}[X|\mathcal{F}_t]=
\begin{cases}
X &\text{ if } X\leq t\\
\mathbb E[X|X>t] &\text{ if } X>t
\end{cases}$$
Here is how I view the conditional expectation. Let $t$ be fixed, $\mathcal F_t$ is a sigma algebra on the base sample space $\Omega$. For each $\omega\in\Omega$ we can write $X(\omega)$ and this will give us a real number, and either $X(\omega)\leq t$ or $X(\omega)> t$. $\mathbb E[X|\mathcal F_t]$ is a random variable still and so given any $\omega\in\Omega$, we can evaluate $\mathbb E[X|\mathcal F_t](\omega)$.
The sigma algebra $\mathcal F_t$ is a collection of subsets of $\Omega$ though. Consider $A\in\mathcal F_t$, then we can ask what $\omega\in A$. If we perform our random experiment and get outcome $\omega$, and if $X(\omega)\leq t$, then $\mathbb E[X|\mathcal F_t](\omega)=X(\omega)$. If we perform our random experiment and get outcome $\omega$ such that $X(\omega)> t$, then $\mathbb E[X|\mathcal F_t](\omega)=\mathbb E[X|X>t]$ which is $X$ averaged over all $\omega$ which give $X$-values above $t$.
Here is another thing, suppose we perform the random experiment and get two different outcomes $\omega_1$ and $\omega_2$ that happen to both give $X$-values above $t$ but with $X(\omega_1)\neq X(\omega_2)$. Then we actually have that $\mathbb E[X|\mathcal F_t](\omega_1)=\mathbb E[X|\mathcal F_t](\omega_2)$. Intuitively, I think of it this way: $\mathbb E[X|\mathcal F_t]$ as a random variable cannot "resolve" any $\omega$ that result in $X$-values above $t$; it sees "$\omega_1=\omega_2$" in a sense because $\omega_1\not=\omega_2$ in reality but it simply can't tell them apart. Like if each $\omega$ has a different color assigned to it, and the each specific color causes $X$ to take some value. Then $\mathbb E[X|\mathcal F_t]$ can tell all colors apart perfectly as long as those colors cause $X\leq t$, but any color that causes $X>t$, it just sees them all as some smeared out average hue.
