I am hoping for guidance in proving the following statement: Given a function $f \in L_p(X)$, for $1 \leq p < \infty$, and a measure $\mu$, then, $$ \lim_{N \to \infty} \mu \left( \lbrace x \in X \ | \ f(x) \leq N \rbrace \right) = \mu(X) \ . $$ This may well have been proven in the literature before, but I am yet to find such a result.
If it makes any difference, I am specifically interested in the case when $X = \Omega$, a sample space with an associated measure $\mathbb{P}: \Omega \mapsto [0,1]$.
Your statement "the set of unbounded $L^p$ functions has measure zero", doesn't make much sense. Note that $\mu$ measures some subsets of $X=\Omega$, not subsets of $L^p$ (a totally different set!). It might be possible defining a proper measure (say $\nu$) on $L^p$ and showing that the set of unbounded functions of $L^p$ is (first of all measurable) and of $\nu$ measure $0$, but that is a totally different thing than what you're describing.
For the limit you want to prove: It is well known that $f\in L^p\implies f $ is a.e. finite. Now let $E_n=\{x: f(x)\leq n\}$. Obviously $E_n\subset E_{n+1}$ and $\displaystyle{\bigcup_{n=1}^{\infty}E_n=X-F}$, where $F$ is of measure zero and that is because $f$ is almost everywhere finite, that is for almost all $x\in X$ there exists $n$ such that $f(x)\leq n$. Now by the "continuity" of measure, $\mu(X)=\mu(X-F)=\mu(\bigcup_{n}E_n)=\displaystyle{\lim_{n\to\infty}\mu(E_n)}$, which is the desired equality.