Suppose that $p(a)\ge 0 $ and $\int_0^{+\infty} p(a)da = 1$. Let $$A= \{ r\in \mathbb{R} \ | \ \forall a\gt 0 \ \ f(r,a) \lt g(r,a) \}$$ and $$B= \{ r\in \mathbb{R} \ | \ \int_0^{+\infty} \exp(f(r,a))p(a)da \lt \int_0^{+\infty} \exp(g(r,a))p(a)da \}$$ Can we conclude that $A = B$?
My try: It's obvious that $A \subset B$ since $$f(r,a) \lt g(r,a) \implies \exp(f(r,a)) \lt \exp(g(r,a)) \implies p(a)\exp(f(r,a)) \lt p(a)\exp(g(r,a)) \\ \implies \int_0^{+\infty} p(a)\exp(f(r,a)) \lt \int_0^{+\infty} \exp(g(r,a))p(a)da$$ I don't know how to show that $B \subset A$: $$\int_0^{+\infty} \exp(f(r,a))p(a)da \lt \int_0^{+\infty} \exp(g(r,a))p(a)da \implies \int_0^{+\infty} [\exp(g(r,a)) - \exp(f(r,a))]p(a)da \gt 0$$ Edit: My main motivation for the question was the case $$f(r,a) = -\frac{(r-6a)^2}{72k} \text{and} \ \ g(r,a) = -\frac{(r-2a)^2}{8k}$$ where $k\gt0$ is a constant.
Counterexample 1: Let $p(a)=e^{-a}$ and $f(r,a)=\sin(a), g(r,a)=\cos(a)$ then obviously $A=\emptyset$ and for every $r\in \mathbb R:$ $$\int_0^\infty e^{\sin(a)-a}\text da< \int_0^\infty e^{\cos(a)-a}\text da\implies B=\mathbb R.$$
Counterexample 2: Its not true even for the monotonic function just use $p$ like before and $f=a/2$ and $g=\sqrt a$.
Note: In general every couple of function $f,g$ such that $f(a_1)>g(a_1)$ and $f(a_2)>g(a_2)$ for some $a_1,a_2\in \mathbb R$ and that exists $\int e^fp\neq \int e^gp$, will produce produce a counterexample $("f(r,a)=f(a),g(r,a)=g(a)")$
For the special case: Lets assume that $p\ge0$ its a (general) function such that $\int p=1$ and the two known integral exists finite for every $r$.Now this functions have this property: $$\lim_{a\to 0}\frac{f(r,a)}{g(r,a)}=\lim_{a\to 0}\frac{(r-6a)^2}{9(r-2a)^2}=\begin{cases}1/9\;\;\text{if } r\neq 0\\ 1\;\;\text{if } r=0\end{cases} $$ and $f(r,a)>g(r,a)$ if and only if:$$ \frac{(r-6a)^2}{9(r-2a)^2}>1 \iff (r-6a)^2>9(r-2a)^2\iff $$ $$36a^2-12ar+r^2>9(4a^2-4ar+r^2)\iff $$ $$24ar>8r^2$$ $\implies \forall r>0$ there exists a certain $a$ such that $f(r,a)>g(r,a)$ and for the previous limit we know that there exists a neighborhood (when $r\neq 0$) such that $f(r,a)<g(r,a)\implies A\subset(-\infty,0]$. But there will definitely be values of $r>0$ such that the two integral are different i.e. $B\cap \mathbb R_+\neq \emptyset$ (for a proper choice of $p$, see "p.s.").
P.S. If you want (for a fixed $r>0$) $\int e^fp<\int e^gp$ just use a function $p(a)$ with support in $[0,\frac{8r^2}{24r}=\frac{r}{3})$ (because here $f<g$). If you want the contrary the $p$'s support has to be $(\frac{r}{3},\infty)$.