Do $L^p$ spaces have the approximation property? A Banach space $X$ has the approximation property if every compact operator $T:X \to X$ is the norm-limit of a sequence of finite-rank operators.
My question is if there is a simple proof that the approximation property holds for $L^p(\Sigma,\mu)$ spaces. It would be enough for me to prove this property in the case that $T$ is linear.
I have found a proof of this in the case $\mu(\Sigma)<\infty$ in which the sequence is explicitely found but I am unable to adapt the proof for the general case.
Any reference would be appreciated too. Thanks
 A: Norbert answered your question but it seems to me that there is no need for such a machinery here.
First, note that if the measure $\mu$ is infinite but $\sigma$-finite, using the Radon–Nikodym theorem, you can find an isometry between $L_p(\mu)$ and $L_p(\nu)$ for some finite measure $\nu$.
Let $p\in [1,\infty)$. Now, suppose that $\mu$ is arbitrary. It is enough to show that each separable subspace of $L_p(\mu)$ is contained in a complemented (separable) subspace with the approximation property. Let $X\subset L_p(\mu)$ be separable. Let $Y$ be the sublattice generated by $X$; clearly $Y$ is separable too. By the Kakutani theorem, $Y$ is complemented and isometric to $L_p(\nu)$ for some $\sigma$-finite measure $\nu$ in which case $Y$ has the approximation property.
For $p=\infty$ consider the pointwise-closed subalgebra generated by $X$.
A: Not even every $L_p$ space for $1\leq p\leq\infty$ have the approximation property, but every $\mathfrak{L}_p^g$ have bounded approximation property. For details see section 23.3 in Tensor Norms and Operator Ideals. A. Defant, K. Floret
