Evaluate the indefinite integral of multiplication of two functions

For $x \in [a,\infty)$, define $F(x) := \int^x_a f(t)\,dt$. Then $\lvert F(x) \rvert \le K$, and $F'(x) = f(x)$ by the fundamental theorem of calculus.

Now for $x,y \in[a,\infty)$, and wlog let $y \ge x$. By integrating by parts and using the triangle inequality, we see \begin{align*}\left\lvert \int^y_a f(t) g(t) \,dt - \int^x_a f(t)g(t)\,dt\right\rvert &= \left \lvert\int^y_x f(t) g(t)\, dt \right \rvert\\ &= \left \lvert \int^y_x F'(t)g(t) \,dt \right \rvert\\ &= \left \lvert [F(t)g(t)]^{t=y}_{t=x} - \int^y_x F(x)g'(x)\,dx \right \rvert\\ &\le \lvert F(y)g(y)\rvert + \lvert F(x)g(x) \rvert + \int^y_x \lvert F(t)g'(t) \rvert \,dt\\ & \le K\left(\lvert g(y) \rvert + \lvert g(x) \rvert + \int^y_x \lvert g'(t)\rvert \,dt \right). \end{align*} Since $g$ is decreasing, we have $g'(t) \le 0$, and so $\lvert g'(t)\rvert = -g'(t).$ Thus \begin{align*}\left\lvert \int^y_a f(t) g(t) \,dt - \int^x_a f(t)g(t)\,dt\right\rvert & \le K\left(\lvert g(y) \rvert + \lvert g(x) \rvert + \int^y_x \lvert g'(t)\rvert \,dt \right)\\ &= K\left(\lvert g(y) \rvert + \lvert g(x) \rvert - \int^y_x g'(t) \,dt \right)\\ &= K(\lvert g(y) \rvert + \vert g(x) \rvert + g(x) - g(y)) \\&\le 4K \text{max}(\lvert g(x) \rvert, \lvert g(y) \rvert). \end{align*} Since $\lvert g(t) \rvert \to 0$ as $t \to \infty$, for any $\epsilon > 0$, there is $M > 0$, such that $\lvert g(t)\rvert < \epsilon/4K$ for all $t > M$. But then for $x,y > M$, we have shown $$\left\lvert \int^y_a f(t) g(t)\, dt - \int^x_a f(t)g(t)\,dt\right\rvert < \epsilon.$$ Finally, take any sequence $\{x_n\}_{n=1}^\infty$ with $x_n \to \infty$, and this shows that $$\left\{\int^{x_n}_a f(t)g(t)\,dt \right\}_{n=1}^\infty$$ is a Cauchy sequence and hence converges. By the sequential criterion theorem, we conclude that $$\lim_{x\to\infty} \int^x_a f(t) g(t) \,dt$$ converges.

As an aside, this is completely analogous to the Dirichlet Test for infinite series, which is a generalization of the alternating series test.