\begin{equation}\int_{-\infty}^{\infty} f \ dx = \frac{1}{2\alpha} \int_{-\infty}^{\infty} \left( \int_{x - \alpha}^{x + \alpha} f \ dx'\right) dx \end{equation}
To show that there is at least one non-trivial solution, consider $f(x) = \frac{1}{1 + x^2}$. We have that $\int_{-\infty}^{\infty} f \ dx = \pi$. We then wish to evaluate
\begin{align*} I = \ &\frac{1}{2\alpha} \int_{-\infty}^{\infty} \left( \tan^{-1}(x + \alpha) - \tan^{-1}(x - \alpha) \right) dx . \ \\ \end{align*}
Note that
\begin{align*} \int \tan^{-1}(u) \ du = u \tan^{-1}(u) - \frac{1}{2} \log(1 + u^2) \ , \end{align*}
and so
\begin{align*} & \ \ \ \ \ \int_{-\infty}^{\infty} \tan^{-1}(x + \alpha) - \tan^{-1}(x - \alpha) \\ =& \lim_{x \to \infty} 2 \left( (x + \alpha) \tan^{-1}(x + \alpha) - (x - \alpha) \tan^{-1}(x - \alpha) - \frac{1}{2} \log\left( \frac{1 + (x + \alpha)^2}{1 + (x - \alpha)^2} \right) \right) \\ =& \lim_{x \to \infty} 2 \left( 2 \alpha \tan^{-1}(x + \alpha) - \frac{1}{2} 0 \right) \\ =& 4 \alpha \frac{\pi}{2} \end{align*}
From which we can conclude that $I = \pi$.
I have found that this also works if $f(x) = e^{-x^2}$, and a few other simple examples where $f$ has compact support. I'm thinking there should be some general result but I'm not sure how to prove it.
If I have well understood your question, I assume that you are using Lebesgue theory. Then your function is $f$ in $L^1(\mathbb{R})$. Put $g(u,x)=f(u)\chi_{[x-\alpha,x+\alpha]}(u)$. We have $$\int_{-\infty}^{+\infty}(\int_{-\infty}^{+\infty}|g(u,x)|dx)du=\int_{-\infty}^{+\infty}2\alpha|f(u)| du<+\infty$$ Hence $g(u,x)$ is integrable on $\mathbb{R}^2$, and you can use Fubini:
$$\int_{-\infty}^{+\infty}(\int_{-\infty}^{+\infty}g(u,x)du)dx=\int_{-\infty}^{+\infty}(\int_{-\infty}^{+\infty}g(u,x)dx)du=\int_{-\infty}^{+\infty}2\alpha f(u) du=2\alpha\int_{-\infty}^{+\infty}f(u)du$$