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In the following theorems:

  1. Theorem. Let $(X,\mathcal{F},\mu)$ be a measure space and $1\le p\le\infty$. For each $f\in L^p(\mu)$ and $\epsilon>0$ there exists a simple function $f_s$ such that $|f_s|\le|f|$ and $||f-f_s||_p<\epsilon$. Moreover, if $p<\infty$, then $f_s$ may be chosen to vanish outside a set of finite measure.
  2. Theorem. Let $1\le p<\infty$, $f\in L^p(\lambda^d)$, and $\epsilon>0$. Then there exists a continuous function $g$ vanishing outside a bouded interval such that $||f-g||_p<\epsilon$.

I do not understand why the assertion in $(1)$ fails for the case $p=\infty$, and also that in $(2)$ does not hold for $p=\infty$.

masacuota
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  • Well, for $p=\infty$ we have counter-examples. For 2. take for example $f \equiv 1$. Then $f \in L^\infty$ but for any such continuous function $g$ we have $|f(x)-g(x)| = 1$ outside the bounded interval. For 1. you might take the characterstic function $\chi_A$ for $A$ constructed in https://math.stackexchange.com/questions/57317/construction-of-a-borel-set-with-positive-but-not-full-measure-in-each-interval – Zahlenteufel Jan 17 '21 at 00:10

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