Improper integral of function $f$, which is Riemann integrable on every closed interval, from $-\infty$ to $\infty$ exists if there exists real number $c$ such that both improper integrals $\int_{-\infty}^{c}f(x) \text{d} x$ and $\int_{c}^{\infty}f(x) \text{d} x$ exist and
$$\int_{-\infty}^{\infty}f(x) \text{d} x=\int_{-\infty}^{c}f(x) \text{d} x+\int_{c}^{\infty}f(x) \text{d} x \text{ .}$$
Is the following definition equivalent to previous definition: improper integral $\int_{-\infty}^{\infty}f(x) \text{d} x=\lim_{\substack{A \to -\infty\\B \to \infty}}\int_A^Bf(x) \text{d} x$? If both improper integrals $\int_{-\infty}^{c}f(x) \text{d} x$ and $\int_{c}^{\infty}f(x) \text{d} x$ exist for some $c$, then $\lim_{\substack{A \to -\infty\\B \to \infty}}\int_A^Bf(x) \text{d} x =\int_{-\infty}^{c}f(x) \text{d} x+\int_{c}^{\infty}f(x) \text{d} x$, but does existence of $\lim_{\substack{A \to -\infty\\B \to \infty}}\int_A^Bf(x) \text{d} x$ imply existence of $\int_{-\infty}^{c}f(x) \text{d} x$ and $\int_{c}^{\infty}f(x) \text{d} x$ for some $c$?
In other words, is the following implication true:
$$((\forall a<b\quad f\in\mathcal R[a,b]) \land(\exists I\in \mathbb R \quad \forall\varepsilon>0 \quad \exists N \in \mathbb N \quad \forall A_1,A_2>N \quad \left \lvert \int_{-A_1}^{A_2}f(x)\text d x -I \right \rvert <\varepsilon)) \implies\\(\exists c\in \mathbb R \quad \exists \int_{-\infty}^{c}f(x) \text{d} x \quad \land \quad \exists\int_{c}^{\infty}f(x) \text{d} x)\text{ ?}$$
My idea was to define a function $F:\mathbb R^2 \mapsto \mathbb R,\quad F(x,y):=\int_x^yf(t)\text d t$ which has the properties:
$$F(x,y)=-F(y,x) \text{,}$$
$$F(x,y)=F(x,z)+F(z,y) \text{.}$$
But I still don't see why would existence of $\lim_{\substack{x \to -\infty \\ y \to \infty}}F(x,y)$ imply existence of $\lim_{x \to -\infty}F(x,z)$ and $\lim_{ y \to \infty}F(z,y)$ for some $z \in \mathbb R$.
Best Answer
Your conjecture is true. Assume $$\lim_{\substack{x \to -\infty \\ y \to \infty}}F(x,y)=I\in\Bbb R$$ and let $\epsilon>0.$ For some $N\in\Bbb N$ we have $$\forall a,b\ge N\quad|F(-a,b)-I|<\epsilon$$ hence $$\forall b,c\ge N\quad |F(0,b)-F(0,c)|=|F(-N,b)-F(-N,c)|<2\epsilon.$$ Therefore, by Cauchy's criterion, $$ \lim_{y\to+\infty}F(0,y)$$ exists (in $\Bbb R$). Similarly (or by difference), so does $$ \lim_{x\to-\infty}F(x,0).$$ Note that the existence of $ \lim_{y\to+\infty}F(z,y)$ and $\lim_{x\to-\infty}F(x,z)$ for some $z\in\Bbb R$ implies their existence for all $z\in\Bbb R.$