Suppose we have an integral like $$ \int_{-\infty}^a f(x)\, dx $$ and want to know if the integral converges as $a\to -\infty$.
Is it correct that this is necessarily converging to $0$, that is $$ \lim_{a\to -\infty}\int_{-\infty}^a f(x)\, dx=0? $$ Intuitively, if $a\to -\infty$, I am integrating over a set of measure $0$.
This depends on the function $f$. If $f(x)=1$ for all $x \in \mathbb R$ then $$\int^a_{-\infty} f(x)dx = +\infty$$for all $a\in \mathbb R$, so the limit is $+\infty$.
However, if you can guarantee that $$\int^A_{-\infty} \lvert f(x)\rvert dx < +\infty$$ for some real number $A$, then yes, your limit will hold true, and it is a consequence of the Monotone convergence theorem or Lebesgue dominated convergence theorem.
EDIT: The proof is as follows. Suppose that for some $A \in \mathbb R$, $$\int^A_{-\infty} \lvert f(x) \rvert dx$$ converges. Then for $a < A$, consider the function $$f_a(x) = \chi_{(-\infty,a]}(x) f(x), \,\,\,\,\, x \in \mathbb (\infty, A],$$ where $\chi_{(-\infty,a]}$ denotes the indicator function of $(-\infty,a]$. This sequence of functions clearly goes to zero pointwise as $a \to -\infty$, and is bounded by the integrable function $\lvert f(x) \rvert$ on the domain $(-\infty, A]$. Thus we can use the LDCT to see \begin{align*}\lim_{a\to-\infty} \int^a_{-\infty} f(x)dx &= \lim_{a\to-\infty} \int^A_{-\infty} f(x)\chi_{(-\infty,a]}(x)dx \\ &= \int^A_{-\infty} \lim_{a\to-\infty} \big( f(x)\chi_{(-\infty,a]}(x) \big) dx \\ &= \int^A_{-\infty} 0 \,\, dx = 0.\end{align*}
Unfortunately, if you can't guarantee any sort of integrability for $f$, this argument (or any similar ones) will not work, and you will need to know something more specific about $f$ to prove the property.
