A condition under which an Lp function is L-infinity I am looking for a condition under which a function in $L_p(\Omega)$ is also in $L_\infty(\Omega)$. The condition may be on the function itself, or on $\Omega$.
In other words, is there anything that guarantees a p-integrable function is bounded?
 A: There is a necessary and sufficient condition. It goes as follows:
Let $f$ be in $L^p$, or even just $L_{loc}^1$. Given a non null measurable subset $A$ of $\Omega$, define the measure space $\mathbf{A} := (A, \mathcal F_A, \nu_A)$ where $\mathcal F_A $ is the induced sigma algebra, and $\nu_A$ is the normalised induced measure given by $\nu_A (E) = \frac{\mu(E \cap A)}{\mu(A)}$.
Given any $f, A$ as above, define $f_A$ to be the function on $\mathbf{A}$ given by $f_A (x) = \frac{ f(x)}{\mu(A)}$.
Let $\mathcal D$ denote the family of non-null $G_{\delta}$ subsets of $\Omega$. Then we have the following:
Theorem: $f$ is in $L^ \infty$ if and only if the family $\{f_A\}_{A \in \mathcal D}$ is uniformly integrable.
Note: The condition that $\mathcal D$ consists of all non-null $G_\delta$ sets can be signifcantly relaxed.
A: You can also use integrability of derivatives, i.e. apply Sobolev embedding: If $\Omega\subset\mathbb{R}^n$, then it holds that the Sobolev space $W^{k,p}(\Omega)$ embeds into $L^\infty(\Omega)$ (even into the continuous and bounded functions), if
$$\tfrac{n}{k}\leq p.$$
In other words: For an $L^p$ function to be bounded, you also need that the derivatives of order up to $n/p$ are also $p$-integrable. For example, in one dimension, $W^{1,1}$ embeds into $L^\infty$, i.e. an integrable function with integrable weak derivative also also bounded.
