Relation between the modulus of integrability and $L^p$ spaces Let $(X,\mu)$ be a   measure space with $\mu(X)<\infty$. Given an integrable function $f$ on $X$, we can quantify its integrability in multiple ways. One is the modulus of integrability, which is a function $\omega:(0,\infty)\to(0,\infty)$ such that 
$$\int_E |f|\,d\mu\le \omega(\mu(E)) $$
for every measurable set $E\subset X$. Another is the $L^p$ space membership: the function $f$ may happen to be in some $L^p(X)$ with $p>1$. 

How are these two concepts related?

This (self-answered) question is motivated by An example of a function not in $L^2$ but such that $\int_{E} f dm\leq \sqrt{m(E)}$ for every set $E$, which is a special case of the above.  
 A: From $L^p$ to $\omega$
If $f\in L^p$ with $1<p\le \infty$, then by Hölder's inequality.
$$\int_E |f| \le \|f\|_p \|\chi_E\|_q = \|f\|_p(\mu(E))^{1/q} $$
where as usual, $p^{-1}+q^{-1}=1$. So, in this case $f$ has the modulus of integrability of the form $\omega(\delta)=C\delta^{1-\frac{1}{p}}$.  This bound is not perfectly tight for $p<\infty$: we can get $\omega(\delta)=o(\delta^{1-\frac{1}{p}})$ by replacing  $\|f\|_p$ with $\|f\chi_E\|_p$ which tends to zero as $\mu(E)\to 0$. However, this $o()$ cannot be quantified without additional information on $f$. 
From $\omega$ to $L^\infty$
If $f$ admits a linear modulus of integrability $ \omega(\delta)=C\delta$, then $f\in L^\infty$. Indeed, let  $E=\{x:|f(x)|\ge C+1\}$ has positive measure:  then 
$$(C+1)\mu(E)\le \int_E |f|\,d\mu\le C\mu(E)$$ hence $\mu (E )=0$.
From $\omega$ to $L^p$, $1<p<\infty$
The function $f(x)=x^{-1/p}$ on the interval $[0,1]$ with the Lebesgue measure $\mu$ fails to be in $L^p$ despite having the modulus of integrability $\omega(\delta)=C\delta^{1-\frac{1}{p}}$. Indeed,
$$\int_E x^{-1/p}\,dx\le \int_0^{\mu(E)}x^{-1/p}\,dx = \frac{p}{p-1}\mu(E)^{1-\frac{1}{p}}$$
while the integral of $|f|^p$ obviously diverges.
A sufficient condition for $f$ to be  in $L^p$ is having a modulus of integrability such that $$\int_0^1 \left(\frac{\omega(t)}{t}\right)^p\,dt<\infty \tag{1}$$
In particular, $\omega(t)=Ct^r$ with $r>1-\frac1p$ suffices, and so does $Ct^{1-\frac1p}/(1+|\log t|)$.
The proof becomes more transparent in terms of the decreasing rearrangement $f^*:(0,\mu(E)]\to[0,\infty)$ defined by $$f^*(t) = \inf\{\lambda : \mu(\{|f|>\lambda\})\le t \}$$
By definition, $f^*$ has the same distribution function as $|f|$, and therefore (by the layercake representation) the same $L^p$ norm. Since $f^*$ is decreasing, 
$$
f^*(t) \le \frac{1}{t}\int_0^t f^*(s)\,ds \le \frac1t \omega(t)
$$
which shows that (1) implies $f\in L^p$. 

For  precise correspondence between a function space and $\omega$ has to consider the Lorentz spaces $L^{p,\infty}$ instead of Lebesgue spaces $L^p$: see, for example, Interpolation of Operators by Bennett and Sharpley.
