Let $\mu$ and $\nu$ be positive measures in $\left(X,\mathrm{\mathcal{M}}\right)$ . Assume that $\mu\left(E\right)\leq\nu\left(E\right)\forall E\in\mathcal{M}$ . Is $L^{4}\left(X,\nu\right)$ contained in $L^{4}\left(X,\mu\right)$ ?
I try to let $f\in L^{4}\left(X,\nu\right)$, I show that $f\in L^{4}\left(X,\mu\right)$. Since $f\in L^{4}\left(X,\nu\right)$, we have $\int_{E}\left|f\right|^{4}d\nu<\infty$. And $\int_{E}\left|f\right|^{4}d\mu\leq\int_{E}\left|f\right|^{4}d\nu<\infty $ (since $\mu\left(E\right)\leq\nu\left(E\right)\forall E\in\mathcal{M}$). That means $f\in L^{4}\left(X,\mu\right)$, so we are done.
Is my proof exact?
Thanks in advanced.
Yes, it's exact. Maybe the inequality $$\int_E g\mathrm d\mu\leqslant \int_E g\mathrm d\nu$$ for $g$ non-negative has to be justified going back to simple functions with non-negative coefficients.