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For the following, consider an arbitrary measure space $(X,\mathcal{M}, \mu)$ In trying to prove that convergence in $L^p$ implies convergence in measure, I came accross this question. Now I'm trying to understand the proof given, I got stuck when it says $$ \Big(\int _X \vert f_n-f\vert ^p\Big)^{1/p} \ge \int_{A_{n,\varepsilon}}\Big(\vert f_n-f\vert ^p\Big)^{1/p} $$ I see why it would be so that $$ \Big(\int _X \vert f_n-f\vert ^p\Big)^{1/p} \ge \Big(\int_{A_{n,\varepsilon}}\vert f_n-f\vert ^p \Big)^{1/p} $$ (exponent outside the integral). Since $A_{k,\varepsilon}\subset X$. Moreover, I also wonder about the more general result given in the question title. Any help apreciated.

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D. Brito
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  • For finite measure spaces, use Jensen's inequality. I think it's likely you want to replace $p$ by $1/p$ throughout... –  Nov 13 '18 at 00:22

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I assume that you mean to ask when

$$\int_X g \ge \left(\int_X g^p\right)^{1/p}$$

holds, since you mentioned $L^p$ norms. Assuming that $g$ is a positive function, this does not hold in very many circumstances. If $X$ has measure $1$, Jensen's inequality implies the reverse inequality and $g$ must be constant on $X$. If $X$ has measure greater than $1$, then apply this argument to measure-1 subsets of $X$. If $X$ has measure less than $1$, then study the measure $\tilde{\mu} := \mu / \mu(X)$.

But for what it's worth, none of that is being used in the linked question, since there's no comparison of $L^1$ and $L^p$ norms. Monotonicity of the $L^p$ norm is the only thing really getting used.