Bounded expectation implied bounded conditional or vice versa? If $\mathrm{E}\left(X\right)<\infty$ does that imply $\mathrm{E}\left(X|Y\right)<\infty$? How about vice versa? 
I'm thinking if we condition on an event (say $Y>2$) then if we have $\mathrm{E}\left(X\right)<\infty$ we can somehow use the theorem of total expectation to bound the other terms. Any thoughts? I've been searching for some kind of list of useful rules about expectations so please let me know if you have such a thing. Thank you!
Edit: After reading what fellow users have said I believe I should ask whether we can bound expectations given events (for example $E(X\mid Y>2)$) if we know that  $\mathrm{E}\left(X\right)<\infty$.
 A: In an attempt to put the language to help us, $E(X\mid Y>2)$ is a "conditional expected value given a specific event", and it is a number, while $E(X\mid Y)$ is shorthand for $E(X\mid \sigma(Y))$ - a conditioning on the whole sigma algebra generated by $Y$, and hence a function of $Y$, called the "conditional expectation of $X$ given $Y$".  
Conditional Expectations are defined for random variables $X$ such that $E(|X|) < \infty$, and they, too, satisfy $E(|E(X\mid Y)|) < \infty$ (this is provable).
Half- formally, the defining property of the function "Conditional Expectation of $X$ given $Y$", write $Z = E(X\mid Y)$ is
$$E(Z\cdot I_{\{B_j\}}) = E(X\cdot I_{\{B_j\}})\;\; \forall B_j $$
where $B_j = \{Y=y_i\}$ is an event (so we have numbers in both side of the equation), and $I_{\{\cdot \}}$ is the indicator function.
Since the indicator function takes the values $0$ or $1$, then  $$E(X\cdot I_{\{B_j\}})=\cases {0 \\  \\
E(X)}$$
Since $E(X) < \infty$, we have $$E(X\cdot I_{\{B_j\}}) < \infty \Rightarrow E(Z\cdot I_{\{B_j\}})< \infty$$
and the known result
$$E(Z) = E\big[E(X\mid Y)\big] = E(X)$$
So, if the Conditional Expectation has a finite expected value, is it possible that itself would be infinite in some sense?
