# $L^1$ and $L^2$ spaces, how to determine which to work with?

This is from a standing point of a new student of measure-theoretic probability. For example, we have the following two definitions of conditional expectation:

Definition 1 ($$L^2$$):

Let $$(\Omega, F , P )$$ be a probability space and let $$G$$ be a $$σ$$−algebra contained in $$F$$ . For any real random variable $$X ∈ L^2(Ω,F,P)$$, define $$E(X |G)$$ to be the orthogonal projection of $$X$$ onto the closed subspace $$L^2(Ω,G,P)$$

Definition 2 ($$L^1$$):

Let $$(Ω,F,P)$$ be a probability space and let $$G$$ be a $$σ$$−algebra contained in F. For any real random variable $$X∈L^1(Ω,F,P)$$, define $$E(X|G)$$ to be the unique random variable $$Z∈L^1(Ω,G,P)$$such that for every bounded, $$G$$−measurable random variable $$Y$$, $$E(XY)=E(ZY)$$.

How do we know which one to work with? And we also hear frequently that "Assume xx is square integrable"..."We require this is in $$L^1$$"..."We require this is in $$L^2$$ etc. But I never undestood, for example, why certain theorems require a random variable to be $$L^1$$-bounded.

Please enlighten me about why this distinction is crucial, either from Probability Theory standing point or Analysis standing point.

• The $L^2$ definition of conditional expectation is easier to visualize and work with, because you have an intuitive understanding of a projection, something that is only available in $L^2$ – rubikscube09 May 4 at 21:02
• In the case of probability spaces, (and all finite measure spaces) $L^p \subset L^q$ if $p > q$, so it is always best in these cases to work with the larges exponent you have. – rubikscube09 May 4 at 21:03