$u\in\mathcal{L^{1}(\mu)}$ (integrable) and $v\in\mathcal{L^{1}(\mu)}\Rightarrow u\cdot v\in\mathcal{L^{1}(\mu)}$ Let $(X,\mathcal{A},\mu )$ be some measure space.

If $u\in\mathcal{L^{1}(\mu)}$ (integrable) and $v\in\mathcal{L^{1}(\mu)}$ (integrable) can I
  then deduce that $u\cdot v\in\mathcal{L^{1}(\mu)}$?

 A: No: take $X=[0,1]$ with Borel $\sigma$-algebra and Lebesgue measure, and $f=g=x^{-1/2}$.
Furthermore, if the claim was true, then $\mathcal L^1(\mu)$ would be contained in $\mathcal L^{2^n}(\mu)$ for each integer $n$. As a consequence of Hölder's inequality, we would have $\cal L^1(\mu)\subset \cal L^p(\mu)$ for any $p>1$. In particular, $L^1\subset L^p$ and the inclusion is continuous by the closed graph theorem: there exists $C_p$ such that $\lVert f\rVert_p\leq C_p\lVert f\rVert_1$ for each $f\in L^1$. If $A$ has finite non-zero measure, then $\mu(A)^{1/p}\leq C_p\mu(A)$ hence $\mu(A)\geq K$ for some positive constant $K$. 
Conversely, if we can find $K>0$ for each set of non-zero measure we have $\mu(A)\geq K$, then your claim is true. Indeed, the sets $\{|u|\geq k\},k\geq 1$ are measurable and form a decreasing sequence. As $u$ is integrable, all these sets have finite measure. Hence for $k$ large enough, $\mu\{|u|\geq k\}<K$ hence have measure $0$. Then we can deduce the claim in the OP is true.  
A: No. For example take $X$ to be the unit circle aroung the origin in $\mathbb{R}^2$ with the Borel $\sigma$-algebra and the Lebesgue measure. Then $u(x)=v(x)=1/|x|$ are each in $\mathcal{L}^1$ but their product is not!
Note, that if you ask $u,v\in\mathcal{L}^2$, then $u\cdot v\in\mathcal{L}^1$ follows by the Hölder inequality.
