Let $D$ be a bounded connected open subset of $R^n$ and $μ$ is a finite measure on $D$, say the Lebesgue measure. Is $L_2(μ)$ separable? Is a bounded sequence $\{f_k\}$ of $L^2(μ)$ pre-compact?
Is $L^2(D)$ separable?
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measure-theory
functional-analysis
hilbert-spaces
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3Yes, $L^2$ is separable (consider step functions or continuous functions with compact support). No, a bounded sequence is not pre-compact: try an orthonormal sequence. – 2012-05-02