The title pretty much says it all: I'm stuck with this problem and I would appreciate any help. I feel like I'm in the middle of nowhere with this one. Let $f:\mathbb{R}\rightarrow \mathbb{R} \in L^1$. For any $t \in \mathbb{R}$ we define the transfer of $f$ by $t$ as the function $f_t:\mathbb{R}\rightarrow \mathbb{R}$ with $f_t(x)=f(x+t),$ for all $x \in \mathbb{R}$. Prove that $\int_{\mathbb{R}}f = \int_{\mathbb{R}}f_t$.
The only thing I have in mind is that if I could somehow prove that $\int_{0}^{\infty}f_t = \int_{t}^{\infty}f$ then in a similar manner I would be able to prove that $\int_{-\infty}^{0}f_t = \int_{-\infty}^{t}f$. I can't see the next step though. I was thinking of defining a sequence $(t_n)$ that converges to $t$, so that the dominated convergence theorem would be accesible, but I can't work it out.
P.S: I would prefer if someone could give me a good hint for this one rather than a detailed solution; any help is appreciated though.
Rather than using Dominated Convergence, can I recommend that you work directly from the definition of the integral?
You could first deal with the case where $f$ is positive. Here, the definition of the integral is $$ \int f(x) dx = \sup_{0 \leq \varphi \leq f, \\ \varphi {\rm \ simple}} \int \varphi(x) dx.$$
But if $\varphi(x)$ is a positive simple function bounded above by $f(x)$, then $\varphi(x + t)$ is a positive simple function bounded above by $f(x + t)$...