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 $p > p\_0$ ?



Total sets for $L^p$ for every $1\leq p < \infty$
No, this is not true. The set $S$ of all simple function orthogonal to $x^{-\frac 13}$ is not dense $L^2(0,1)$. To see that it is dense in $L^1(0,1)$, take $s$ simple, let $A=\int\_0^1 s(x)x^{-\frac 13}$ and choose $t=\frac 23 A \delta^{-\frac 23}\chi\_{(0, \delta)}$. Then $s-t \in S$ and $\|t\|\_1 \leq C \delta^{\frac 13} \to 0$ as $\delta \to 0$.


