Let $f$ a piecewise continous function with period $2\pi$,and $\mathcal{P}=C_{per}^{\infty}[-\pi,\pi]$(the space of infinitely differentiable functions $g:\mathbb{R}\rightarrow\mathbb{C}$ periodics,with period $2\pi$).
A linear functional on $\mathcal{P}$ ,$T:\mathcal{P}\rightarrow \mathbb{C}$,is called a periodic distribution if there exists a sequence $(\psi_{n})_{n\geq 1}$in $\mathcal{P}$ such that $$\langle T,\varphi\rangle=T(\varphi)=\lim_{n\to\infty}\int\limits_{-\pi}^{\pi}\psi_{n}(x)\varphi(x)dx,\;\forall\varphi\in\mathcal{P}.$$ The question is: We want to prove that $T_{f}:\mathcal{P}\rightarrow \mathbb{C} $ given by $$\langle T_{f},\varphi\rangle=T_{f}(\varphi)=\int\limits_{-\pi}^{\pi}f(x)\varphi(x)dx,$$is a periodic distribution.
Since the definition of $piecewise$ $continuous$ is not always that clear I will addionally assume that $f\in L^1[-\pi,\pi]$, otherwise $T_f(1)\not \in \mathbb C$ with $1$ as the constant function. The key to that problem is the fact that the smooth functions are dense in the $L^p$ spaces. Therefore we can find some sequence $f_n$ such that $\int_{-\pi}^\pi |f(x)-f_n(x)| dx$ tends to zero as $n$ increases. With that we get \begin{align*} \left | \int_{-\pi}^\pi f(x)\varphi(x) dx-\int_{-\pi}^\pi f_n(x)\varphi(x)dx \right |&=\left | \int_{-\pi}^\pi(f(x)-f_n(x))\varphi(x) dx \right|\\ &\leq \int_{-\pi}^\pi |f(x)-f_n(x)|\cdot |\varphi(x)| dx\\ &\leq ||\varphi||_\infty \int_{-\pi}^\pi|f(x)-f_n(x)| dx\\ &\longrightarrow 0. \end{align*} In other words we have $T(\varphi)=\lim_{n\to \infty} \int_{-\pi}^\pi f_n(x)\varphi(x)dx$.