The space $L^{\infty -}$ and showing it is an algebra. Let 
$$\mathcal{A}=L^{\infty -}(\Omega,P):=\bigcap_{1\le p<\infty}L^p(\Omega,P)$$
I've already showed it is a vector space over $\mathbb{C}$. So, with the usual multiplication of (complex-valued) functions operation I want to prove that

$fg\in\mathcal{A}\quad\forall f,g\in\mathcal{A}\\\text{Which is the same as to prove that }\forall f,g\in\mathcal{A}\\
\int_{\Omega}|fg|^p\;dP<\infty,\quad 1\le p<\infty$

But got stucked. I only get that $fg\in L^1(\Omega,P)$ thanks to Hölder's inequality:
$$
\int_{\Omega}|fg|\;dP\le \Big(\int_{\Omega}|f|^p\,dP\Big)^{1/p}\Big( \int_{\Omega} |g|^q\;dP \Big)^{1/q},\quad\frac{1}{p}+\frac{1}{q}=1\quad 1\le p,q< \infty
$$
for $p=q=2$. From here I can't see how $fg\in L^p(\Omega,P)$ for $p\ge2$. Any help or hints would be appreciated.

Robert's answer is really great. My answer is along the lines of what you're doing already. Let $r \ge 1$, then
$$\int_{\Omega} |fg|^r \,dP = \int_{\Omega} |f^r||g^r|\,dP.$$
Define auxiliary functions $\widetilde{f} = |f^r|$ and $\widetilde{g} = |g^r|$, then
$$\int_{\Omega} |fg|^r\,dP = \int_{\Omega} |\widetilde{f}\widetilde{g}|\,dP.$$
Use Holder on this to get that
$$\int_{\Omega} |fg|^r\,dP \le \left(\int_{\Omega} |\widetilde{f}|^p\,dP\right)^{1/p} \left(\int_{\Omega} |\widetilde{g}|^q\,dP\right)^{1/q} = \left(\int_{\Omega} |f|^{pr}\,dP \right)^{1/p} \left(\int_{\Omega} |g|^{qr}\,dP \right)^{1/q}.$$
Here $p$ is chosen such that the integrals are finite, but.. you already know that $f$ and $g$ are both in every $L^p$ space, so these are automatically finite.
