Let $\Omega=(0,\infty)$ and $\mathcal F=\mathcal{B}(\Omega)$. Let $\mathbb P$ be the probability measure corresponding to the exponential distribution with parameter $\lambda$. $X$ and $Y$ are two random variables given by $X(\omega)=\omega$ and $Y(\omega)=\min(\omega,k)$ for some $k>0$. In this setting I have to find $E(X\mid \sigma(Y))$.
I have $\sigma(Y)=\{\phi,\Omega,\{\omega\ge k\},\{\omega\le k\}\}$. Defining $Z=E(X\mid \sigma(Y))$, $Z(\omega)=X(\omega)$ if $\omega\le k$. If $\omega\ge k$, $Z(\omega)=\int_k^\infty x \lambda e^{-\lambda x} \mathrm dx$.
While I think the above answer is correct, I am unable to use the formal definition of conditional expectation to arrive at the answer. The formal definition simply says $Z$ is $\sigma(Y)$-measurable $\int ZdP=\int XdP$, where the integrals are taken over any set in $\sigma(Y)$. How to draw the connection between the definition and these results?
For every integrable $X$ and every nonnegative $x$, $$ E(X\mid\min\{X,x\})=X\cdot\mathbf 1_{X\lt x}+E(X\mid X\geqslant x)\cdot\mathbf 1_{X\geqslant x}. $$ Edit: The sigma algebra generated by $\min\{X,x\}$ is the collection of events $\{X\in B\}$ where $B$ is in $\mathcal B(\mathbb R)$ and, either $B\subseteq[0,x)$ or $[x,+\infty)\subseteq B$.
These correspond to events $\{\min\{X,x\}\in A\}$ where $A$ is in $\mathcal B(\mathbb R)$ and, either $A$ does not contain $x$ or $A$ contains $x$, respectively.