If $f$ is Lebesgue integrable, does $$\int_{\mathbb R}|f|=\lim_{t\to \infty }=\int_{-t}^t |f|\ \ ?$$
This question may be obvious and stupid, but in an exercise, I have to prove that $$\int_{\mathbb R}|f(x-h)|dx=\int_{\mathbb R}|f(x)|dx,$$ and in the solution, they prove the statement for caracteristic function, then simple function, then positive measurable function then for measurable function. But why don't we simply do :
$$\int_{-t}^t |f(x-h)|dx=\int_{-t-h}^{t-h}|f(x)|dx,$$and taking $t\to \infty $, we get $$\int_{\mathbb R}|f(x-h)|dx=\int_{\mathbb R}|f(x)|dx.$$
Does this work or not really ?
True. $I_{(-t,t)} |f| \to |f|$ pointwise and $I_{(-t,t)} |f| \leq |f|$ and you can apply DCT. You can also use Monotone Convergence Theorem.