In Lars Hörmander's Linear Partial Differential Operators (1963 edition) the following is stated on the bottom of page 3.
... Further, which is important for the definition of distributions, a measure is uniquely determined by its restriction to $ C_0^\infty (\Omega) $. For if $ \mathrm{d} \mu $ is a measure in $ \Omega $ and $ \int u \: \mathrm{d} \mu = 0 $ for every $ u \in C_0^\infty (\Omega) $, then $ \int u \: \mathrm{d} \mu = \lim \int u_\epsilon \: \mathrm{d} \mu = 0 $ for every $ u \in C_0^0 (\Omega) $.
I would be very grateful if someone would be so kind to help me understand this comment. What does it mean for a measure to be "determined completely"? And why does the second sentence demonstrate the foregoing assertion? I am a little stuck...
N.B. $ u_\epsilon $ was previously defined by $ u_\epsilon(x) = \int u(x-\epsilon y) \: \varphi (y) \: \mathrm{d}y = \epsilon^{-n} \int u(y) \: \varphi(x-y/\epsilon) \: \mathrm{d}y $, where $ \varphi \in C_0^\infty (\mathbb{R}^n) $ with $ \mathrm{supp} (\varphi) = \{ x : \lvert x \rvert \leq 1 \} $, $ \varphi \geq 0 $ and $ \int \varphi \mathrm{d} x = 1 $ (i.e. $ \varphi $ is a mollifier).
"Determined completely" means that if $\int f\,d\nu_1 = \int f\,d\nu_2$ for every $f \in C^\infty_0(U)$, then $\nu_1 = \nu_2$.
To see why the second sentence is true, let's suppose $\mu$ is a finite measure. We know that $u_\epsilon$ converges to $u$ pointwise. Moreover, if $C = \sup |u|$ (which is finite because $u$ is continuous and compactly supported), you can easily check that $|u_\epsilon| \le C$ as well. Hence $\int u_\epsilon\,d\mu \to \int u\,d\mu$ by the dominated convergence theorem (using $C$ as your dominating function).
Now if it holds for finite measures, then it also holds for signed measures $\mu$ (consider the Jordan decomposition $\mu = \mu^+ - \mu^-$).
Now suppose $\nu_1, \nu_2$ are finite measures and $\int f\,d\nu_1 = \int f\,d\nu_2$ for all $f \in C^\infty_0$. Let $\mu = \nu_1 - \nu_2$. Then $\int f\,d\mu = 0$ for all $f \in C^\infty_0$. As discussed above, it follows that $\int f\,d\mu = 0$ for all $f \in C^0_0$. This means $\mu$ must be the zero measure; this is the easy direction of the Riesz-Markov theorem, and can be proved using the monotone class theorem or some variant. If $\mu = 0$ then $\nu_1 = \nu_2$.
If $\nu_1, \nu_2$ are not necessarily finite, but are finite on compact sets, then you get the same conclusion by approximating $\Omega$ by a sequence of bounded sets.