Rewrite Lebesgue integral over a centered interval

Question

Given a function $f\in L^1(\mathbb{R})$, define the integral over an interval centered at $x$: $$f_c(x)=\frac{1}{2c}\int_{x-c}^{x+c} f(t)\,dt$$

There is a stated result that we can rewrite $f_c(x)$ into $$f_c(x)=\int_{\mathbb{R}}f(x-t)\frac{1}{2c}\mathbf{1}_{[-c,c]}\,dt$$where $\mathbb{1}_{[-c,c]}$ is the indicator function of $[-c,c]$.

But it seems to be counter-intuitive and I don't know how to proceed. Any idea how to show this is true?

Thanks in advance.

Answer 1

\begin{eqnarray} f_c(x) &=& \int {1 \over 2c} 1_{[x-c,x+c]}(t) f(t) dt \\ &=& \int {1 \over 2c} 1_{[-c,+c]}(t-x) f(t) dt \\ &=& \int {1 \over 2c} 1_{[-c,+c]}(x-t) f(t) dt \\ &=& \int {1 \over 2c} 1_{[-c,+c]}(\tau) f(x-\tau) dt \\ \end{eqnarray}

Answer 2

Using the definition of the indicator function, $$ \int_{\mathbb{R}} f(x-t) \frac{1}{2c} 1_{[-c,c]} \,dt = \frac{1}{2c} \int_{-c}^{c} f(x-t) \,dt. $$ Now just make a change of variables $s=x-t$.