Do measures only have meaning in reference to some space? Does it only make sense to take the measure (whatever measure that may be) whilst considering the space it is in? So for example in $\mathbb{R}$, $[0,1]$ would have some non-zero measure but in $\mathbb{R}^2$ it would have a measure of zero since it occupies no "volume" (area) in this space?
Or have I misunderstood what a measure is actually meant to represent?
 A: See Hausdorff measure.
What you are noticing is a very fundamental concept related to Hausdoff measure. Namely, $[0,1]$ has measure $1$ as measured with the Lebesgue measure on $\mathbb{R}^1$, but if you embed $[0,1]$ into $\mathbb{R}^n$ for $n>1$, then it has Lebesgue measure zero as measured by the Lebesgue measure on $\mathbb{R}^n$.
This may be a little confusing, but it is no contradiction. The problem here is that although we say "the" Lebesgue measure, it is actually a different measure for each $n$.
But there certainly is something to it. Clearly $[0,1] \times\{0\}$ has no volume in $\mathbb{R}^2$, but doesn't it still have length $1$? The answer is yes it does, and we can formalize this thinking with the Hausdorff measure. Informally speaking, the $d$-dimensional Hausdorff measure $\mathcal{H}^d$ in $\mathbb{R}^n$ measures the $d$-dimensional volume of an object in $\mathbb{R}^n$. For instance, in $\mathbb{R}^3$ this means $\mathcal{H}^1$ measures length, $\mathcal{H}^2$ measures surface area, and $\mathcal{H}^3$ measures usual volume in $\mathbb{R}^3$.
For example, if $\gamma:[0,1]\to\mathbb{R}^3$ is a $C^1$ not self intersecting curve in $\mathbb{R}^3$, then
$$
\mathcal{H}^1(\mathrm{range}(\gamma)) = \int_0^1|\gamma'(t)|dt
$$
which agrees with our intuition of length of a curve. Moreover, if $d=n$ we have $\mathcal{H}^n$ agrees with the $n$-dimensional Lebesgue measure.
The Hausdorff measure goes even further though, it is defined for all $d \geq 0$, not just $d \in \mathbb{N}$. This lets us compute the "dimension" of objects that, for instance, are too big to have finite length, but too small to have positive surface area. For examples, many fractals fall into this category of object. This is called the fractal dimension (or Hausdorff dimension) of the object.
A: Yes! By definition, a measure is a non-negative map on certain subsets of a space. These subsets must form a $\sigma$-algebra by definition, so in particular, the complement of a subset must also be "measurable". In your example, considering $\mathbb R \subset \mathbb R^2$, we would have that the complement of $[0,1]$ would have to be measurable. Thus, a measure $\mu$ on $\mathbb R$ can't generally be extended to $\mathbb R^2$.
A: Yes, you're right, the concept of a 'measure' is dependent on what the space is. 
It's similar to how answer to the question "Can we factor $f(x) = x^2 + 1$?" depends on what you're factoring $f(x)$ over. If it's the rationals, or the reals, then the answer is no. But over the complex numbers, it does factor. 
Or, in group theory, how it doesn't make sense to ask if a group is normal; we can only say whether a group is normal when considered as a subgroup of another group; normality means nothing, without respect to another group. A 'measure' in a vacuum makes no sense, we need a space whose subsets we'd like to measure, and the result may very well change when the space changes.
