About non-measurable set If there is any non-measurable set $X$, then do we say that it is non-measurable with respect to a particular measure or is it non-measurable in general?
For e.g., we clearly have examples of non-measurable sets on real line. Which means that if $(\mathbb{R}, \mathcal{A})$ is a measurable space then a non-measurable set $X$(say) will not belong to $\mathcal{A}$. However in my Probability Theory class, prof gave us an example of counting measure that is defined on $(\mathbb{R}, 2^{\mathbb{R}})$ where $2^{\mathbb{R}}$ is the power set of $\mathbb{R}$. Now since a power set is the collection of all subsets of the underlying set, it will contain the set $X$ too, which suggests that $X$ is  non-measurable with respect to a some given measure. However, I am not quite sure what actually is the case.
This is my first introduction to Measure theory and that too as a part of the Probability Theory course, so it will be of help if the answer can be put in simple words.
 A: The set-up is that we have a set $(X,\mathcal{A})$ where $\mathcal{A}$ is a $\sigma$-algebra on $X$ (a measurable space, analogous to a topological space). Subsets of $X$ that are in $\mathcal{A}$ are called measurable (like open in topology), subsets that are not in $\mathcal{A}$ are non-measurable (we could call those sets non-open in a topology, though this is not a usually considered thing).
So in the case of the power set all subsets are measurable by definition. And all functions on it are measurable too. Not very interesting. In many practical cases we both have a topology and a $\sigma$-algebra on a set and often related (Borel sets, Baire sets, etc.) Then it gets more interesting IMHO..
If you add the function $\mu: \mathcal{A} \to \overline{\mathbb{R}}$ obeying the right axioms, we say we have a "measure space" (not merely measurable, but we then have fixed/chosen a measure on it too). Then we can talk about sets that have a finite measure, or are measurable for the outer measure etc. (see later in a course on real analysis). 
So it's not the measure that tells us if a set is measurable but the $\sigma$-algebra on which it is defined.
