If $f,g$ each have property p then $f \circ g$ also has property $p$ What type of characteristics should properties have for the following to hold true?
If $f,g$ each have property $p$ then $f  \circ g$ also has property $p$
Examples:
If $f,g$ continuous then $f  \circ g$ is continuous
If $f,g$ entire then $f  \circ g$ is entire
If $f,g$ differentiable then $f  \circ g$ is differentiable 
If $f,g$ one to one then $f  \circ g$ is one to one 
If $f,g$ Contractible then $f  \circ g$ is Contractible
If $f,g$ polynomial then $f  \circ g$ is polynomial
If $f,g$ lineaer then $f  \circ g$ is linear
Counter Examples:
If $f,g$ integrable then $f  \circ g$ is integrable (false) see Robert Israel's answer 
If $f,g$ measureable then $f  \circ g$ is measureable
Update Edit:
The motivation for the question was to use the statement as a filter to specify properties p, I had seen too often question being asked for a specific property, was wondering if there is a way to devise a test for p rather than test each p individually to see if it satisfies the statement. (This is as good as I can explain my intent for asking the question, making a list seems to be a beneficial side effect). If anyone can elucidate the motivation with better mathematical terminology please edit. 
 A: "If $f, g$ integrable then $f \circ g$ is integrable." - False!
For example, $f(x) = g(x) = e^{-x}$ is integrable on $(0,\infty)$, but
$f(g(x)) = e^{-e^{-x}}$ is not integrable there.
A: When all identity maps satisfy your property, then this is, at heart, a sub-category, where the objects are the same and we take a subset of the morphisms. 
You probably want to confine yourself to cases where all identity maps satisfies your property, because the cases where some identity functions are not in the set are "odder."
For example:


*

*Functions which are not onto.

*Functions which are not 1-1.

*Functions which are contractible to a point.

*Functions that factor through a class of sets - $f:A\to B$ factoring though the class if there is a $C$ in the class such that $f$ factors as $A\to C\to B$ for some functions.

*The case of constant functions is a special case of 4.


These cases also have a side-property that give $f,g$, for $g\circ f$ to have the property, you need only one of them to have the property. For (1), you only need $g$ in the class. For (2), if $f$ is in the class. For (3-5), If either $f$ or $g$ satisfies our property, then $g\circ f$ is.
So, these cases are sort of like "ideals" in the parent category, rather than sub-categories.
The cases (1) and (2) act like prime ideals - if $g\circ f$ is not one-to-one, then one of $f$ or $g$ is not one-to-one.

If you have a collection of these properties (either with the identities or not,) then the intersection of them is also a property like this. So differentiable one-to-one functions are closed under composition. 
