Is this a measurable space? This probably has a very quick answer but it's been bugging me for a while.
Take two copies of the Cartesian plane. On each plane cut out the square $S=(0,1)\times(0,1)$. Throw away the rest. Now make a cut in each square from the middle of the top edge $(1/2,1)$ to the center $(1/2,1/2)$. Identify the left edge of the cut on one square with the right edge of the cut on the other square, and vice versa. That's my space $X$. My question is: is this a "measure space"? I.e. is the measure just inherited from my original Euclidean measure on the planes or can the cut get me into trouble?
The context of my question is that I have an integral kernel $k(x,y)$ defined on $X\times X$ and I want to study the integral operator $K$ associated with it. In most of the literature I could find on integral operators, $X$ is assumed to be either a subset of the Euclidean space (say the Wikipedia article or Stone's book) or, more generally, a measure space. I want to use the theorems on integral operators (in particular the fact that when $K$ is Hilbert-Schmidt, then the $L^2(X\times X)$-norm of $k(x,y)$ is equal to the sum of the squared eigenvalues of $K$). So I just want to make sure that my $X$ satisfies the necessary conditions for those theorems.
 A: If you would like to talk about integrals, are you talking about the measurable space, perhaps? A measure space is a measurable space on which a measure is fixed. A measurable space can admit many measures. 
Now, your initial space is $X' :=[0,1]^2\times\{0,1\}$ which comes with a Borel topological and measurability structure inherited from $\Bbb R^2$. If I understood your "identifying" operations correctly, you introduce a quotient map $f$ between topological spaces $X'$ and $X$. More precisely, you introduce a topology on $X$ using $X'$ and $f$. For such topology you can talk about Borel measurability, which is a quite natural structure here, which is indeed inherited from $\Bbb R^2$. In general, whenever you consider $X$ as being a topological space, no matter how you constructed it, you can always introduce Borel $\sigma$-algebra on $X$. 
Also, if you are primarily interested in measures, for any measure $\mu$ on $X'$ you can define an image/pushforward  measure $f_*\mu$ on $X$ using the very same quotient map $f$. The good thing about such measure is that you can integrate w.r.t. it on $X'$ rather than on $X$ using the change of variables formula.
A: Let $M'$  be the space you define, then $M := M' \setminus \{0\}$ is a differentiable (actually it's $C^\omega$) manifold as it is an open subset of the riemann surface associated to $x\mapsto \sqrt[2]{x}$ (minus the branching point at $0$).
Let then $p$ be the projection $M\to \Bbb R^2\setminus\{0\}$, and let $\cal U$ be a cover by simply connected open sets of $M$ so that $p(U)\cong U$. For each $U\in \cal U$ let $\mu_U$ be a measure on $M$ defined as $\mu_U(A) = m(p(A\cap U))$. Now take a partition of unity $\lambda_U$ associated to $\cal U$ and define $\tilde\mu (A)= \sum_U (\int_{A\cap U}\lambda_U d\mu_U)$ and finally define the measure $\mu$ on $M'$ as $\mu(A) = \mu(A\cap M)$.
