Condition on $f$ in $L^{p, \infty} $ implies $f \in L^q$ If $f \in L^{p, \infty}$ and $\mu ( \{ x : f(x) \neq 0 \} ) < \infty$, then $f \in L^q$ where $q <p$. 
$\textbf{My Attempt:}$ 
Let $E = \{x:f(x) \neq 0 \}$
$$ || f ||_q^q = q \int_{0}^{\infty} \alpha^{q-1} df(\alpha) d \alpha$$ 
$$ = q \int_{0}^{\infty} \alpha^{q-1} \int_{ \{ x: |f(x)| > \alpha \}} 1 dx$$
$$ \le q \int_{0}^{\infty} \alpha^{q-1} d \alpha \int_{ E \cap \{x : |f(x)| > \alpha \} } 1 dx$$
Usually, I would use Fubini's theorem and make a change of variables such that $ \alpha < \frac{1}{|f(x)|}$ but to do this I need that $x \in E$ . Any help on how to deal with the intersection in the integrand??

Let me give this a shot.  So $f\in L^{p,\infty}$ means that 
$$
\mu(\{|f(x)|>t\}) \leq Ct^{-p}
$$
for a constant $C$ independent of $t$.  This is useful for large $t$.  We also 
know that $\mu(\{|f|>0\}) < \infty$, so for another constant we have 
$$
\mu(\{|f(x)|>t\}) \leq D,
$$
and this is useful for small $t$.
Like you suggested, we'll use the layer-cake formula to estimate $\|f\|_q$.  We can split up the integration over $\alpha$ into two pieces, according to which of the above bounds we prefer to use.
\begin{align*}
\|f\|_q^q &= q \int_0^\infty \alpha^{q-1} \mu(\{|f|>\alpha\})\,d\alpha \\
&\leq Dq \int_0^1 \alpha^{q-1}\,d\alpha + Cq\int_1^\infty \alpha^{q-p-1}\,d\alpha.
\end{align*}
Both integrals are bounded, the second because the power $q-p-1 < -1$.  If you want a sharp bound, you would optimize the point at which you switch between the two estimates (I arbitrarily chose $1$) based on which inequality is better for a given $t$.  This is fine to show it's finite though.
