Does $(\alpha=\int_{-\infty}^{\infty} f(x)\, \mathrm{d}x)$ follow from $(\lim_{k\to \infty} \int_{a_k}^{b_k} f(x)\,\mathrm{d}x=\alpha)$?

Today I learned a bit about integrals and I'd like to know if that is true, or if there is a counterexample that disproves the following:

When an $$(\alpha\in \mathbb{R})$$ exists, so that for all sequences $$((a_k)_{k\in \mathbb{N}}, (b_k)_{k\in \mathbb{N}} \subset\mathbb{R})$$ with $$(a_k\to -\infty, b_k\to \infty)$$ for $$(k\to \infty)$$ already $$(\lim_{k\to \infty} \int_{a_k}^{b_k} f(x)\,\mathrm{d}x=\alpha)$$ applies, then the following $$(\alpha=\int_{-\infty}^{\infty} f(x)\, \mathrm{d}x)$$ applies too.

My thoughts are that when $$(a_k\to -\infty, b_k\to \infty)$$ for $$(k\to \infty)$$ applies, then at some time (meaning for $$(k\geq k_0)$$ with a $$(k_0\in \mathbb{N}))$$ the following: $$(a_k<0)$$ and $$(b_k>0)$$ hold.

So the integral can be split like that: $$(\int_{a_k}^{b_k} f(x)\mathrm{d}x=\int_{a_k}^{0} f(x)\mathrm{d}x+\int_{0}^{b_k} f(x)\mathrm{d}x)$$ which leads us to the definition of improper integrals.

Is my argumentation correct or wrong? Because from $$(\lim_{n\to \infty} (c_n+d_n))$$ the existence of $$(\lim_{n\to \infty}c_n)$$ or $$(\lim_{n\to \infty}d_n))$$ do not follow...

$$\lim_{k\rightarrow\infty}\int_{c_k}^{d_k}f=0$$
for any two sequences $$c_k,d_k\rightarrow\infty$$, or equivalently
$$\lim_{x,y\rightarrow\infty}\int_x^yf=0$$
Hence By Cauchy $$\lim_{x\rightarrow\infty}\int_0^xf$$ converges. Similarly $$\lim_{x\rightarrow-\infty}\int_x^0f$$ converges so overall $$f$$ is integrable over $$\mathbb{R}$$. Choosing $$a_k=-k,\ b_k=k$$ gives you the integral is $$\alpha$$.