Consider $L^p[ 0,1]$ for $1\leq p < \infty$ or, if you prefer, $L^p(\mu)$ where $\mu$ is a finite Borel measure with compact support. Let $(\phi)_{i\in I}$ be a subset of measurable functions that is contained in $L^p$ for every $1\leq p < \infty$ and assume that it is total for $L^{p_0}$ for some $1\leq p_0< \infty. $
Could you deduce that $(\phi_i)_{i\in I}$ is total for $L^p$ for every $1\leq p < \infty$? If $1\leq p_1 \leq p_0$ it is enough to recall that the set of linear combinations of step functions is dense in every $L^p$, and argue as it follows. Let $f \in L^{p_1}$ and $\varepsilon > 0.$ Then, there exists $g \in L^{p_1}\cap L^{p_0}$, which is a linear combination of step functions, such that $||f-g||_{p_1} < \varepsilon.$ Moreover, since $(\phi)_{i\in I}$ is total in $L^{p_0},$ there exists $h$ in its linear span such that $||g-h||_{p_0} < \varepsilon.$ Since the measure is finite, it follows that $||g-h||_{p_1} \leq C ||g-h||_{p_0},$ with $C > 0.$ Finally, $$||f-h||_{p_1} \leq ||f-g||_{p_1} + ||g_h||_{p_1} < \varepsilon (1+C),$$ as we wanted. What does it happen with the rest of exponents $p > p_0$ ?
See the first answer here.
The important thing they show is that if $0\neq f\in L^1[0,1]\setminus L^2[0,1]$ (they work in $\mathbb{R}^n$, but the argument works in our setting as well), then there exists an orthonormal basis $(\phi_i)_{i\in \mathbb{N}}$ of $L^2[0,1]$, with $\phi_i\in C^\infty[0,1]$, and such that $$ \int_0^1 \varphi_i(x)f(x)\, dx =0, \qquad i=1,2,\ldots $$ On the other hand, recall that a subspace $H$ in a Banach space $X$ is total if and only if $\{ \ell\in X^*: \ell(x)=0, \, \forall x\in H\} =\{0\}$. If we use $X=L^p[0,1]$ with $1\leq p<\infty$, then $X^*=L^{p'}[0,1]$ ($1/p+1/p'=1$) by the Riesz representation theorem.
Combining these two sets of results we obtain, by fixing $1<q<2$ and choosing $f\in L^q\setminus L^2$, an example of a sequence $(\phi_i)_{i\in \mathbb{N}}$ that is in every $L^p$, total in $L^2$, but not total in $L^{q'}$.