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It is well-known that (given a measure space $(S,\mathcal A,\mu)$ and $1\le p\le\infty$) the Banach space $L^p(S,\mathcal A,\mu)$ has infinite dimension.

Is there an easy way to proof this statement (or a suitable reference (preferably a book) where I can find this result)?

Martin Sleziak
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gifty
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    if $\mathcal{A}$ is finite, you can't prove it. Otherwise consider indicator functions of all measurable sets – Norbert Aug 09 '12 at 11:15
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    You can find necessary and sufficient conditions for finite-dimensionality here (in case $\mu$ is totally finite --- remember that $L^p \subset L^1$ under this assumption). @Norbert: it also depends on the measure... – t.b. Aug 09 '12 at 11:38
  • You are also in trouble if $\mu = 0$. – ncmathsadist Aug 09 '12 at 11:39

2 Answers2

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As Norbert mentioned, this is simply not true if $S$ is finite.

In general, suppose you can find a collection of countably infinite pairwise disjoint measurable sets $\{ A_n:n\in \mathbb{N}\}$, each with finite positive measure. Then, the collection $\{ 1_{A_n}:n\in \mathbb{N}\}$ is an infinite linearly independent set contained in $L^p$. Why? First of all, the fact that each $A_k$ has finite measure guarantees that $1_{A_k}$ is an element of $L^p$. As for linear independence, suppose we have some finite linear combination of these functions that is equal to $0$: $$ a_11_{A_{n_1}}+\cdots +a_m1_{A_{n_m}}=0. $$ Now, multiply this equation by $1_{A_{n_k}}$ and integrate. You will find that $$ a_k\mu (A_{n_k})=0, $$ and hence, because $\mu (A_{n_k})>0$, we have that $a_k=0$, which proves linear independence.

Jonathan Gleason
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Since $l^{p}$ embeds isomorphically into $L^{p}$,and it's easy to check that $l^{p}$ is infinitive dimensional.

Tomas
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