Showing some Function $fg$ is Integrable when $g(x)\le e^{-\vert x\vert }$ and given some Conditions on $f$. Suppose $\int_a^b \vert f \vert <\infty$ for all real $a,b$ and that $$\int_{-r}^r \vert f \vert \le (r+1)^a$$ for all real $r$ some real $a$, and that $$g(x)\le e^{-\vert x \vert}$$ I want to show $$f(x)g(tx)\in L^1(\mathbb{R})$$ when $t\not =0$.
My biggest issue is finding a decent upper bound for $\vert f \vert$. My idea is to find an upper bound for $\vert f \vert$ on $[-r,r]$, use this upper bound to find an upper bound for $\vert f(x)g(tx)\vert$ such that when I integrate from $-r$ to $r$ and take a limit as $r\to \infty$ I have convergence. The problem is that although the integral of $f$ is bounded on $[-r,r]$, $f$ can assume arbitrarily large values over this interval. I blindly used $(r+1)^a$ as an upper bound for $\vert f\vert$ but when I do that I get $\int _{-r}^r \vert f(x)g(tx) \vert \le \int _{-r}^r (r+1)^a\cdot e^{-\vert xt\vert}\le  (r+1)^a(\frac{e^{tr}-e^{-tr}}{t})$ if $t>0$, but this tends to $\infty$ as $r\to \infty.$
How can I find a sharper upper bound for $\vert f\vert$ or show that $fg\in L^1(\mathbb{R})$?

Motivation for this solution: we're given information about the integral of $f$, so we want to bring that function into the bounds somehow. That suggests integration by parts.
Let's just work on $[0,\infty)$ (the other half is similar) and assume $t>0$. Define $F(x) = \int_0^x |f(y)|\,dy$. Then
$$
\int_0^r |f(x)g(tx)|\,dx \le \int_0^r |f(x)| e^{-tx}\,dx = F(x)e^{-tx}\big]_0^r + t \int_0^r F(x) e^{-tx}\,dx.
$$
Since $F(0)=0$ and $F(x) \le (x+1)^a$, this gives
$$
\int_0^r |f(x)g(tx)|\,dx \le (r+1)^ae^{-tr} + t \int_0^r (x+1)^ae^{-tx}\,dx;
$$
taking $r\to\infty$ yields
$$
\int_0^\infty |f(x)g(tx)|\,dx \le t \int_0^\infty (x+1)^ae^{-tx}\,dx,
$$
which is a convergent integral.
