How does one integrate with respect to a surface measure, is it the same as a surface integral? I've tried a google search but couldn't find much.
In particular I have the following problem but can't begin as i don't know the above!
Let $\Gamma \in L^{1}(S^{n-1})$ where $S^{n-1}$ is the (n-1) dimensional sphere on $\mathbb{R}^{n}.$ Given $z \in \mathbb{R}^{n} \backslash \{0\}$ the projection onto $S^{n-1}$ is given by: $$ z' = \frac{z}{|z|} \in S^{n-1}$$
We define the operator $$Tf(x) = \lim_{\epsilon \to 0}\int_{|z| > \epsilon}\frac{\Gamma(z')}{|z|^{n}}f(x-z)dz.$$
Show that the limit above exists for $f \in \mathcal{S}(\mathbb{R}^{n})$ i.e. a Schwartz function by assuming that $\int_{S^{n-1}}\Gamma(z')d \sigma(z')=0$ where $d \sigma$ denotes the surface measure on $S^{n-1}$.
You should have a look in Stein's Singular Integrals and Differentiability Properties of Functions from 1970, in Chapter II, Section 4.2., Theorem 3.
To answer your initial question while being a bit nitpicky, it is epistemologically very strange to say that "surface measure" is the same as "surface integral". As you probably learned in a measure theory course, measures are set functions (they take as input sets), whereas the integral (which has to be defined with respect to a measure) can be regarded as a linear operator defined on various function spaces.
However, I guess that your question is rather: "Is computing some integral with respect to surface measure the same as computing surface integrals in calculus?". In this case, the answer is yes.
For instance, if you have two space dimensions and take $\Gamma(x,y)=\max\{x,0\}$ for $(x,y)\in S^1$, i.e., $x^2+y^2=1$, then you can just use polar coordinates on the circle. If I'm not mistaken, the integral $\mathrm{d}\sigma$ equals 2 in this case.