Alex, when reading the title it looks like you ask for conditions to inclusion between $L^p$-spaces, but in your query, it looks more like you are into the convergence of Fourier series. Since these matters are different I split the answer into two parts.
Inclusion between $L^p$-spaces
When the measure space, i.e. the domain of integration, is
Finite (like $[0,1]$), then as @timur pointed out, we have $$L^p\subset L^q, \quad \text{provided $p>q>0$} $$
Infinite and non-discrete, then as @timur pointed out there is no inclusion.
Infinite and discrete, then
$$L^p\subset L^q, \quad \text{provided $q>p>0$} $$
see How do you show monotonicity of the $\ell^p$ norms?
These matters are connected to the Fourier series in a theorem known as the Pontryagin duality. The interval $[0,1]$ can be viewed upon as the circle group, and the dual group of the circle group is the group of integers (a discrete infinite group) - in case of the real numbers, the dual group is the real numbers again.
Convergence of Fourier series
Fourier series are series of the form $f(x)\sim\sum_{n=-\infty}^\infty c_ne^{2\pi inx}$,
with $c_n=\int_0^1f(x)e^{-2\pi inx}dx$.
Such series are defined for $f\in L^p[0,1]$ when $p\geq1$. When talking about convergence of Fourier series one usually means pointwise convergence (almost everywhere since sets of zero measures are unimportant in the Lebesgue sense).
Kolmogorov constructed a function in $L^1$ where the Fourier series, which do exist, diverges almost everywhere (1922).
Note that Kolmogorov's example shows that extra conditions (like the Dini-test) upon $L^1$-functions to have convergent Fourier series are non-superfluous.
For long it was an open problem if the Fourier series of a $L^2$-function converge or not, and it was proved to be true by Carleson in 1962, and was later sharpened to $L^p$ $1<p\leq2$ by Hunt.
Interestingly, it could be mentioned that there are intermediate spaces $A$ and $B$, with $L^1\supset A\supset B\supset L^p$ for all $p>1$, where $A$ contain functions with divergent Fourier series and while $B$ does not which has been studied by Konyagin and others.