under which conditions this equality holds Consider $f : [0,\infty) \rightarrow \mathbb{R}$ be a function such that $\lim_{t\rightarrow \infty} f(t) = 0$. I was wondering if the following relation holds
$$lim_{t\rightarrow\infty}\int_0^t |f(\tau)|d\tau = lim_{t\rightarrow\infty}\int_\lambda^t |f(\tau-\lambda)|d\tau$$ for some constant $\lambda>0$. Intuitively I'm thinking it is reasonable as guess, since if $f$ tends to zero and $\lambda$ if finite, as $t$ grows to infinity then, very informally, $f([t-\lambda,t])$ tends not to contribute to the left integral. However I'm looking for a quite rigorous proof for that (if it exists) and also to some conditions $f$ must satisfy.
What I thought was to rewrite the first integral as
$$ \int_0^t |f(\tau)|d\tau = \int_\lambda^{t+\lambda} |f(\xi - \lambda)|d\xi = \int_\lambda^t|f(\xi -\lambda)|d\xi + \int_t^{t+\lambda}|f(\xi-\lambda)|d\xi$$
and then to show the the two latter integral does not contribute as $t\rightarrow\infty$, however I don't know how to deal with them. For example i can write, for $t\ge \lambda$
$$\int_t^{t+\lambda}|f(\xi-\lambda)|d\xi = \int_0^{\lambda}|f(\chi -\lambda+t)|d\chi$$ and the trying to show that $$lim_{t\rightarrow\infty}\int_0^{\lambda}|f(\chi -\lambda+t)|d\chi = \int_0^{\lambda}lim_{t\rightarrow\infty}|f(\chi -\lambda+t)|d\chi = 0$$ however I don't know what conditions I should pose for $f$ in such a way to be able to do that. I looked at the Lebesgue dominated convergence theorem, however I'm not dealing with successions of functions. Any hint?
 A: Ok, let's start with
$$ \int_0^t |f(\tau)|d\tau = \int_\lambda^{t+\lambda} |f(\xi - \lambda)|d\xi = \int_\lambda^t|f(\xi -\lambda)|d\xi + \int_t^{t+\lambda}|f(\xi-\lambda)|d\xi$$
Then $$\exists t_1: \forall t>t_1: f(t-\lambda)<\frac{1}{\lambda} \Rightarrow \int_t^{t+\lambda}|f(\xi-\lambda)|d\xi < 1$$ and $$\exists t_2: \forall t>t_2: f(t-\lambda)<\frac{1}{\lambda^2} \Rightarrow \int_t^{t+\lambda}|f(\xi-\lambda)|d\xi < \frac{1}{\lambda}$$ and so forth $$\exists t_n: \forall t>t_n: f(t-\lambda)<\frac{1}{\lambda^n} \Rightarrow \int_t^{t+\lambda}|f(\xi-\lambda)|d\xi < \frac{1}{\lambda^{n-1}}$$
This sequence converges obviously for $\lambda > 1$ and your guess is correct.
The other cases shouldn't be too hard.
A: You can see that with the new equation you propose you have:
$\int_\lambda^t |f(\tau-\lambda)|d\tau=\int_0^{t-\lambda} |f(u)|du$
So  $\int_0^t |f(\tau)|d\tau-\int_\lambda^t |f(\tau-\lambda)|d\tau=\int_{t-\lambda}^t |f(\tau)|d\tau$
You use the definition of $\lim_{t\rightarrow \infty} f(t) = 0$
That is $\forall \epsilon,\lambda >0$ , $\exists t_0 \forall t>t_0-\lambda$, $|f(t)|<\dfrac{\epsilon}{\lambda}$
etc.
