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I'm trying to calculate $||f||_{L_{\infty}(E)}$ where $$ f(x) = \left\{\begin{matrix} & x, x \in \mathbb{I}\\ & 2x, x \in \mathbb{Q} \end{matrix}\right.\: \: \: E:= [0,1] $$

I knew that $E$ had finite measure. Hence I can use following $$ \lim_{p \to \infty}||f||_{L_{p}(E)} = ||f||_{L_{\infty}(E)} $$

My attempt:

$$ ||f||_{L_{\infty}([0,1])} = \lim_{p \to \infty}\left(\int_{[0,1]}|f(x)|^{p}dx\right)^{\frac{1}{p}} = \lim_{p \to \infty}\left(\int_{[0,1]\cap\mathbb{Q}}|2x|^pdx + \int_{[0,1]\cap\mathbb{I}} |x|^{p}\right)^{\frac{1}{p}} = \lim_{p \to \infty }\left(\int_{[0,1]\cap\mathbb{Q}}(2x)^{p} + \int_{[0,1]\cap\mathbb{I}} x^{p} \right)^\frac{1}{p}=\lim_{p \to \infty}\left(\frac{2^p + 1}{p + 1} \right)^\frac{1}{p} = 2 $$ Am I wrong? If yes: where am I wrong? How could I solve it?

user717043
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1 Answers1

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The rationals have Lebesgue measure zero, so $\Bbb{Q} \cap [0,1]$ has measure zero and so you have incorrectly evaluated one integral: $$ \int_{[0,1] \cap \Bbb{Q}} \dots = 0 \text{.} $$ Correct that and your argument is repaired.

A different argument: $f$ is bounded by $1$ on $[0,1] \cap \Bbb{I}$ (a set having measure $1$) and $f$ is bounded by $2$ on $[0,1] \cap \Bbb{Q}$ (a set having measure $0$). Therefore, $f$ is essentially bounded (see also essential supremum of $|f|$) by the former bound, $1$, $||f||_{\infty} = 1$.

The reasoning is the same in both cases -- the bound on the measure zero set does not alter the essential bound.

Eric Towers
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