Thereom: Let $(X,S,\mu)$ be a measure space. If $\mu(X)<\infty$, $1\leq p < q <\infty$, then $L_q\subset L_p$.

But if we take $X=(0,1]$ and take the Lebesgue measure $\mu$, and define $f(x):=\frac{1}{x^{1/3}}$. Then clearly $\mu(X)=1<\infty$, and $f\in L_2$ but $f\notin L_3$, which contradicts the theorem above. But I can't see what I'm missing.

  • $\begingroup$ Indeed, $L_3\subsetneq L_2$, so $L_2\setminus L_3\neq \emptyset$. You exhibited an element of that set. $\endgroup$ – Lorenzo Quarisa Jun 16 '18 at 15:06
  • $\begingroup$ @LorenzoQ. But then the theorem would be incorrect? $\endgroup$ – Sid Caroline Jun 16 '18 at 15:52
  • 1
    $\begingroup$ No: the theorem implies that $L_3\subset L_2$. This is not a contradiction with $L_3\subsetneq L_2$, which is what you showed. $\endgroup$ – Lorenzo Quarisa Jun 16 '18 at 15:57

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