Can we think of the Fourier series as a Bochner integral?

$\newcommand{\norm}[1]{\|#1\|}$ $\newcommand{\vp}{\varphi}$ $\newcommand{\Z}{\mathbb Z}$ $\newcommand{\T}{\mathbf T}$ $\newcommand{\C}{\mathbf C}$

Let $\T$ denote the unit circle and $\lambda$ denote the normalized Haar measure on $\T$. For each $n\in \Z$ let $\chi_n:\T\to \C$ be the function which takes $z$ to $z^n$.

For any $f\in L^2(\T, \lambda)$ we know that $$ f = \sum_{n\in \Z} \hat f(n) \chi_n $$ in $L^2(\T, \lambda)$. This means that $$ \norm{f - \sum_{n=-N}^N \hat f(n)\chi_n}_2 \to 0 $$ as $N\to \infty$.

Question. My question is can we think of the sum $\sum_{n\in \Z} \hat f(n) \chi_n$ as a Bochner integral?

Here is the natural thing to go for. Let $c$ be the counting measure on $\Z$ and let $\vp:\Z\to L^2(\T, \lambda)$ be the function defined by $$ \vp(n) = \hat f(n) \chi_n $$ If $\vp$ were Bochner integrable then the integral $\int_\Z \vp\ dc$ exists and is nothing but $\sum_{n\in \Z} \hat f(n) \chi_n$ and things are fine. However, the Bochner integrability criterion asks for the finiteness of $\int_\Z \norm{\vp}_2\ dc$, which is same as asking for the finiteness of $$ \sum_{n\in \Z} |\hat f(n)| $$ But since $\hat f$ is not necessarily in $\ell^1(\Z)$, we cannot say that $\vp$ is Bochner integrable.

In general, I would like to think of any infinite sum as an integral under the counting measure of a suitable function. The fact that I am unable to do this for the Fourier series is a bit disconcerting. Perhaps I am making a mistake somewhere?

$𝑓$ seems to satisfy the requirements for being the 'vector-valued integral' of $\int \phi(n)dn = \int \hat{f}(n)\chi(n)dn$ as given in defn 3.26 of Rudin's functional analysis. Namely, for each linear functional $\Lambda$ on $L^2(T)$, $\Lambda\circ \phi$ is integrable and, moreover, $\int \Lambda \circ \phi(n)dn = \Lambda(f)$. I don't know how this relates to Bochner integrability.