I am trying to understand the proof for the Poisson summation formula for $\phi \in \mathcal{S}(\mathbb{R^n})$ i.e. functions in the Schwartz space.
(A1) Now, we want to first show,
$$ g(x) := \sum_{m \in \mathbb{Z}^n} \phi(x + 2\pi m) = (2\pi)^{-n}\sum_{m \in \mathbb{Z}^n} \hat{\phi}(m)e^{ix\cdot m} =: \tilde{g}(x)$$
To show this, we first showed that they have the same Fourier coefficients, i.e. $g_k = \int e^{-it\cdot k}g(t) dt= \int e^{-it\cdot k}\tilde{g}(t) dt=\tilde{g}_k$.
(A2) However, we now need to show that same Fourier coefficients imply identity of function.
The book uses this lemma
I understand that the lemma says that if all Fourier coefficients of a measure are $0$, then the measure is identically 0.
(A3) My professor explained the application to the lemma as follows: knowing the identity of Fourier coefficients for $g$ and $\tilde{g}$, we let $\mu = g-\tilde{g}$ and apply the previous lemma to conclude that $g - \tilde{g} = 0 \implies g = \tilde{g}$ identically.
My question
I wasn't sure how constructing $\mu$ as such leads to a Borel measure. Is the measure given by $\int_{(a,b)} d\mu(x) := [g(b)-\tilde{g}(b)] - [g(a) - \tilde{g}(a)]? $
For any integrable function $h\colon X\to \mathbb C$ on a topological measure space $(X,dx)$, we get a finite (complex-valued) Borel measure $d\mu$ on $X$ defined by $$ d\mu = h\,dx. $$ Super concretely, the measure is defined on Borel sets $E$ by the formula $$ \mu(E) = \int_Eh\,dx. $$ You should check that this formula satisfies the definition of a finite complex-valued measure by using the usual convergence theorems of Lebesgue integration.
In your application, we set $d\mu = (g-\tilde g)\,dx$, where $dx$ is the Lebesgue measure on $\mathbb T^n$. In case $n = 1$, the correct formula you are imagining would involve the derivatives $g',\tilde g'$ on the left-hand side, rather than $g,\tilde g$ themselves.