We know that the average of $f$ over $[x-T,x+T]$ for any fixed $x$ is given by $$f_{\text{avg}}(x)=\frac{1}{2T}\int_{x-T}^{x+T} f(t)\,\mathrm{d}t.$$ Define $\phi=\frac{\Pi_T}{2T}$, where \begin{align}\Pi_T(t)=\begin{cases}1,&\quad -T\leq t\leq T\\
0,&\quad |t|>T.\end{cases}\end{align} We then have $$\int_{-\infty}^\infty \phi(t)\,\mathrm{d}t=\frac{1}{2T}\int_{-T}^T \Pi_T(t)\,\mathrm{d}t=1.$$ For any fixed $x$, \begin{align}\phi(x-t)=\begin{cases}\frac{1}{2T},&\quad x-T\leq t\leq x+T\\
0,&\quad \text{else}.\end{cases}\end{align} Thus, for any given function $f$ and any fixed value of $x$, \begin{align}f(t)\phi(x-t)=\begin{cases}\frac{1}{2T}f(t),&\quad x-T\leq t\leq x+T\\
0,&\quad \text{else}.\end{cases}\end{align} We conclude that $$f_{\text{avg}}(x)=\int_{-\infty}^\infty f(t)\phi(x-t)\,\mathrm{d}t.$$ More generally, when we replace $\phi$ with an arbitrary function, we have the definition of a convolution: the convolution of two functions $f$ and $g$ is given by $$f*g(x)=\int_{-\infty}^\infty f(t)g(x-t)\,\mathrm{d}t.$$ Reference: The Mathematics of Imaging by Timothy G. Freeman
