Absolute integrability and limits Suppose that $h \in L_1(\mathbb{R})$ and define
$$
g(t) = \int_{t}^{\infty}h(\tau)\,\mathrm{d}\tau. 
$$
Is it true that $\lim_{t\to\infty}g(t) = 0$? Seems 'obviously' true to me, but what's the formal way to argue it?
 A: Let's first prove it with the assumption that $h\geq 0$. Consider the function $h_t(\tau) = h(\tau)\chi_{[0,t]}(\tau)$, where $\chi$ is the indicator function. Clearly $h_t\leq h$ and $h_t\to h$ pointwise as $t\to\infty$. Thus the dominated convergence theorem implies
$$\lim_{t\to\infty}\int_0^t h(\tau)\,d\tau = \int_0^\infty h(\tau)\,d\tau.$$
Now, since $h\in L^1(\mathbb{R})$ we can rearrange to find that
$$g(t) = \int_t^\infty h(\tau)\, d\tau = \int_0^\infty h(\tau)\,d\tau - \int_0^t h(\tau)\,d\tau.$$
Taking the limit $t\to\infty$ above proves the claim.
Now for general $h$ we can decompose it into positive and negative parts: $h =h^+ - h^-$ where $h^+,h^-\geq 0$ and both are $L^1$. Linearity of integration and limits now gives us the result in this case.
One word of caution: technically the dominated convergence theorem requires us to use a sequence of functions, rather than a continuously parameterized family like $h_t$. On the other hand, checking a limit $t\to\infty$ is equivalent to checking it for all sequences $t_n\to\infty$, so we can move between the two without issue.
