How to explain the existence of this integral? This is a theorem from Theory of Functions of a Real Variable by Natanson.

Let a Lebesgue integrable function ${\large f(t)}$ possessing the property
$$
  {\large M= \sup_{0\le h\le b-a}  {\left |\frac{1}{h} \int_{a}^{a+h}f(t)dt \right|} < + \infty \\ }
  $$
be defined on the closed interval ${\large [a,b]}$.
For any nonnegative decreasing function ${\large g(t)}$ ( the case where ${\large g(a)= +\infty }$ is not excluded ) , defined and Lebesgue integrable on ${\large [a,b]}$, the integral
$$
  {\large \int_{a}^{b} f(t)g(t)dt}
  $$
exists and the inequality
$$
  {\large \left |\int_{a}^{b} f(t)g(t)dt\right|\le M\int_{a}^{b} g(t)dt }
  $$
is valid.

The author proved 
$$
  \left | \int\limits_{\gamma }^{\beta } f(t)g(t)dt \right | \le 
M\left \{ \int\limits_{a}^{\gamma } g(t)dt + \int\limits_{a}^{\beta } g(t)dt \right \} \left ( a < \gamma < \beta < b \right ) 
  $$
and said that the integral ${\large \int_{a}^{b} f(t)g(t)dt}$ exists because
$${\large \lim_{\beta \to a} \int\limits_{\gamma }^{\beta } f(t)g(t)dt= 0}$$
But I cannot understand why it implies the existence.
The conclusion is trivial when the function ${\large f(t)}$ is nonnegative. But it's not a hypothesis. And we can't treat the limit 
$${\large \lim_{\gamma  \to a} \int\limits_{\gamma }^{b} f(t)g(t)dt}$$
as the value of the integral directly.
I have been thinking for a long time, but I still don't know what to do.
part1
part2
part3
 A: I think I see how to do this.
The idea is that the bound that the book proves gives the following property: For every $\varepsilon>0$ there exists $\delta>0$ such that if $0<\beta<\gamma<\delta$, then
$$
\left| \int_\beta^\gamma f(t)g(t)\, dt \right|= \left| \int_{\beta}^b f(t)g(t)\, dt - \int_{\gamma}^b f(t)g(t)\, dt \right|<\varepsilon.
$$
Note that both of the integrals above exist, since $g$ is bounded away from $a$ (owing to it being integrable, nonnegative, and decreasing).
This means that the "continuous sequence" $\left(\int_\beta^b f(t)g(t)\, dt\right)_{0<\beta\leq b}$ is Cauchy. In particular you can check that this implies that for any sequence $\beta_n\to a$ from above, the limit exists and is independent of the particular sequence. This unique limit is what we call $\int_a^b f(t)g(t)\, dt$, and then the bound the book claims indeed follows from letting the parameters $\beta,\gamma$ tend to the endpoints.
Edit: To see that this need not be a usual Lebesgue integral, we can set $f(x)= \frac{\sin(1/x)}{x^\alpha}$, and $g(x)= \frac{1}{x^{1-\alpha}}$ for some $0<\alpha\leq 1$ and $a=0$, $b=1$ (to see that $f$ satisfies the hypotheses in the OP, do the change of variables $x=1/y$ and then integrate by parts). Then
$$
\int_\beta ^1 f(x)g(x)\, dx = \int_\beta^1 \dfrac{\sin(1/x)}{x}\, dx = \int_1^{1/\beta} \dfrac{\sin(x)}{x}\, dx,
$$
which has a limit as $\beta\to 0^+$, but the function $\sin(x)/x$ is not Lebesgue integrable on $[1,\infty)$.
