Properties of improper integral (showing that: $\int \limits_{0}^{\infty}f(x)dx=\int \limits_{0}^{1}f(x)dx+\int \limits_{1}^{\infty}f(x)dx.$) Let $f(x)$ is integrable on every segment $[r,\infty)$ where $r>0$. Let $\int \limits_{0}^{1}f(x)dx$ and $\int \limits_{1}^{\infty}f(x)dx$ converges. Why in this case we can conclude that $$\int \limits_{0}^{\infty}f(x)dx=\int \limits_{0}^{1}f(x)dx+\int \limits_{1}^{\infty}f(x)dx.$$
Can anyone explain it rigorously?
Maybe I don't know a definition.
 A: I think it should help to know that, given an integrable function $f$, it can be said that for $a<b$ and $c \in (a,b)$
$$\int_{a}^{b} f(x)dx = \int_{a}^{c} f(x)dx + \int_{c}^{b} f(x)dx$$
So then of course we can then just take $a=0$, $c=1<b$.
$$\int_{0}^{b} f(x)dx=\int_{0}^{1} f(x)dx+\int_{1}^{b} f(x)dx$$
And let $b \to \infty$.
Proof of this property is trivial. Via the fundamental theory of calculus we know that
$$\int_{a}^{b} f(x)dx=F(b)-F(a), F(x)=\int f(x)dx$$
Therefore we have
$$\int_{a}^{c} f(x)dx + \int_{c}^{b} f(x)dx=F(c)-F(a)+F(b)-F(c)=F(b)-F(a)$$
$$=\int_{a}^{b} f(x)dx$$
A: I think is the case that you "don't know a definition" (as you said) because there is nothing to prove here (at least, in Tom Apostol's approach).
If the integrals
$$\int \limits_{0}^{c}f(x)dx\quad \text{and}\quad \int \limits_{c}^{\infty}f(x)dx\tag{1}$$
are both convergent for some $c\in(0,\infty)$, then
$$\int_0^\infty f(x)dx=\int \limits_{0}^{c}f(x)dx+ \int \limits_{c}^{\infty}f(x)dx$$
by definition. What you can prove if you want is "that the choice of $c$ is unimportant" (Apostol's words), that is, the value of $\int_0^\infty f(x)dx$ doesn't depends on $c$ (provided that both integrals in $(1)$ are convergent).
In your case, 
$$\int \limits_{0}^{1}f(x)dx\quad \text{and}\quad \int \limits_{1}^{\infty}f(x)dx$$
are both convergent and thus
$$\int_0^\infty f(x)dx=\int \limits_{0}^{1}f(x)dx+ \int \limits_{1}^{\infty}f(x)dx$$
by definition of $\int_0^\infty$.
As I said, this answer follows the Apostol's approach (see section 10.23).
A: For any fixed $D$  such that $[D,\infty)\subset$ dom$(f)$  define $$(1)....\quad \int_D^{\infty}f(x) dx=\lim_{Y\to \infty}\int_D^Y f(x) dx   $$ provided that this limit exists. . If $\int_A^{\infty}f(x) dx$  exists for fixed $A$ then for fixed $B>A$  let $$\int_A^Bf(x) dx= C.$$ Then $C$ is fixed and $$(2)....\quad\int_A^{\infty}f(x) dx=\lim_{Y\to \infty}\int_A^Y f(x) dx=\lim_{Y\to \infty} \left ( \int_A^B f(x) dx+\int_B^Y f(x) dx\right )=$$ $$=\lim_{Y\to \infty}\left (C+\int_B^Y f(x) dx \right )=\left (C+\lim_{Y\to \infty}\int_B^Y f(x) dx\right )= \left (\int_A^B f(x) dx+\int _B^{\infty} f(x) dx\right ).$$The first equality in (2) is by (1) with $D=A$, and the last equality, above, is by (1) with $D=B.$ In particular let $A=0,\; B=1.$
