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Tomasz Kania
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As for the first question the answer is no even if $\mu$ is assumed to be $\sigma$-finite as by a simple change of measure you may reduce the problem to the case where $\mu$ is finite. Let us fix a finite measure $\mu$.

When $p=1$ we may use the following reasoning. The space $L_1(\mu)$ is weakly compactly generated because the inclusion map ${\rm id}\colon L_2(\mu)\to L_1(\mu)$ has dense range. On the other hand, it is well-known that $\ell_1(\Gamma)$ does not embed into a weakly compactly generated space for any uncountable set $\Gamma$, however by the principle of local reflexivity you will find it in some ultrapower of $L_1[0,1]$.

When $p\in (1,2)$, then $\ell_1(\Gamma)$$\ell_p(\Gamma)$ does not embed into $L_p(\mu)$ with $\mu$ finite for $\Gamma$ uncountable; this is proved in:

P. Enflo, H. P. Rosenthal, Some results concerning $L_p(\mu)$-spaces. J. Functional Analysis 14 (1973), 325–348.

This was extended to $p\in (2,\infty)$ by Bill Johnson and Gideon Schechtman (see this paper).

As for the second question, should such construction exist, it is most likely covered in one of Fremlin's Measure theory volumes.

As for the first question the answer is no even if $\mu$ is assumed to be $\sigma$-finite as by a simple change of measure you may reduce the problem to the case where $\mu$ is finite. Let us fix a finite measure $\mu$.

When $p=1$ we may use the following reasoning. The space $L_1(\mu)$ is weakly compactly generated because the inclusion map ${\rm id}\colon L_2(\mu)\to L_1(\mu)$ has dense range. On the other hand, it is well-known that $\ell_1(\Gamma)$ does not embed into a weakly compactly generated space for any uncountable set $\Gamma$, however by the principle of local reflexivity you will find it in some ultrapower of $L_1[0,1]$.

When $p\in (1,2)$, then $\ell_1(\Gamma)$ does not embed into $L_p(\mu)$ with $\mu$ finite for $\Gamma$ uncountable; this is proved in:

P. Enflo, H. P. Rosenthal, Some results concerning $L_p(\mu)$-spaces. J. Functional Analysis 14 (1973), 325–348.

This was extended to $p\in (2,\infty)$ by Bill Johnson and Gideon Schechtman (see this paper).

As for the second question, should such construction exist, it is most likely covered in one of Fremlin's Measure theory volumes.

As for the first question the answer is no even if $\mu$ is assumed to be $\sigma$-finite as by a simple change of measure you may reduce the problem to the case where $\mu$ is finite. Let us fix a finite measure $\mu$.

When $p=1$ we may use the following reasoning. The space $L_1(\mu)$ is weakly compactly generated because the inclusion map ${\rm id}\colon L_2(\mu)\to L_1(\mu)$ has dense range. On the other hand, it is well-known that $\ell_1(\Gamma)$ does not embed into a weakly compactly generated space for any uncountable set $\Gamma$, however by the principle of local reflexivity you will find it in some ultrapower of $L_1[0,1]$.

When $p\in (1,2)$, then $\ell_p(\Gamma)$ does not embed into $L_p(\mu)$ with $\mu$ finite for $\Gamma$ uncountable; this is proved in:

P. Enflo, H. P. Rosenthal, Some results concerning $L_p(\mu)$-spaces. J. Functional Analysis 14 (1973), 325–348.

This was extended to $p\in (2,\infty)$ by Bill Johnson and Gideon Schechtman (see this paper).

As for the second question, should such construction exist, it is most likely covered in one of Fremlin's Measure theory volumes.

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Tomasz Kania
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As for the first question the answer is nono even if $\mu$ is assumed to be $\sigma$-finite as by a simple change of measure you may reduce the problem to the case where $\mu$ is finite. Let us fix a finite measure $\mu$.

When $p=1$ we may use the following reasoning. If $\mu$ is $\sigma$-finite, then theThe space $L_1(\mu)$ is weakly compactly generated. Indeed, by a simple change of measure you may assume that $\mu$ is finite and then because the inclusion map $L_2(\mu)\to L_1(\mu)$${\rm id}\colon L_2(\mu)\to L_1(\mu)$ has dense range. On the other hand, it is well-known that $\ell_1(\Gamma)$ does not embed into a weakly compactly generated space for any uncountable set $\Gamma$, however by the principle of local reflexivity you will find it in some ultrapower of $L_1[0,1]$.

When $p\in (1,2)$, then $\ell_1(\Gamma)$ does not embed into $L_p(\mu)$ with $\mu$ finite for $\Gamma$ uncountable; this is proved in:

P. Enflo, H. P. Rosenthal, Some results concerning $L_p(\mu)$-spaces. J. Functional Analysis 14 (1973), 325–348.

This was extended to $p\in (2,\infty)$ by Bill Johnson and Gideon Schechtman (see this paper).

As for the second question, if there isshould such a construction exist, it should beis most likely covered in one of Fremlin's Measure theory volumes.

As for the first question the answer is no.

When $p=1$ we may use the following reasoning. If $\mu$ is $\sigma$-finite, then the space $L_1(\mu)$ is weakly compactly generated. Indeed, by a simple change of measure you may assume that $\mu$ is finite and then the inclusion $L_2(\mu)\to L_1(\mu)$ has dense range. On the other hand, it is well-known that $\ell_1(\Gamma)$ does not embed into a weakly compactly generated space for any uncountable set $\Gamma$, however by the principle of local reflexivity you will find it in some ultrapower of $L_1[0,1]$.

When $p\in (1,2)$, then $\ell_1(\Gamma)$ does not embed into $L_p(\mu)$ with $\mu$ finite for $\Gamma$ uncountable; this is proved in:

P. Enflo, H. P. Rosenthal, Some results concerning $L_p(\mu)$-spaces. J. Functional Analysis 14 (1973), 325–348.

This was extended to $p\in (2,\infty)$ by Bill Johnson and Gideon Schechtman (see this paper).

As for the second question, if there is such a construction, it should be covered in one of Fremlin's Measure theory volumes.

As for the first question the answer is no even if $\mu$ is assumed to be $\sigma$-finite as by a simple change of measure you may reduce the problem to the case where $\mu$ is finite. Let us fix a finite measure $\mu$.

When $p=1$ we may use the following reasoning. The space $L_1(\mu)$ is weakly compactly generated because the inclusion map ${\rm id}\colon L_2(\mu)\to L_1(\mu)$ has dense range. On the other hand, it is well-known that $\ell_1(\Gamma)$ does not embed into a weakly compactly generated space for any uncountable set $\Gamma$, however by the principle of local reflexivity you will find it in some ultrapower of $L_1[0,1]$.

When $p\in (1,2)$, then $\ell_1(\Gamma)$ does not embed into $L_p(\mu)$ with $\mu$ finite for $\Gamma$ uncountable; this is proved in:

P. Enflo, H. P. Rosenthal, Some results concerning $L_p(\mu)$-spaces. J. Functional Analysis 14 (1973), 325–348.

This was extended to $p\in (2,\infty)$ by Bill Johnson and Gideon Schechtman (see this paper).

As for the second question, should such construction exist, it is most likely covered in one of Fremlin's Measure theory volumes.

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Tomasz Kania
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As for the first question the answer is no.

When $p=1$ we may use the following reasoning. If $\mu$ is $\sigma$-finite, then the space $L_1(\mu)$ is weakly compactly generated. Indeed, by a simple change of measure you may assume that $\mu$ is finite and then the inclusion $L_2(\mu)\to L_1(\mu)$ has dense range. On the other hand, it is well-known that $\ell_1(\Gamma)$ does not embed into a weakly compactly generated space for any uncountable set $\Gamma$, however by the principle of local reflexivity you will find it in some ultrapower of $L_1[0,1]$.

When $p\in (1,2)$, then $\ell_p(\omega_1)$$\ell_1(\Gamma)$ does not embed into $L_p(\mu)$ with $\mu$ finite

P. Enflo, H. P. Rosenthal, Some results concerning $L_p(\mu)$-spaces. J. Functional Analysis 14 for (1973), 325–348.$\Gamma$ uncountable; this is proved in:

P. Enflo, H. P. Rosenthal, Some results concerning $L_p(\mu)$-spaces. J. Functional Analysis 14 (1973), 325–348.

This was extended to $p\in (2,\infty)$ by Bill JohnsonBill Johnson and Gideon Schechtman (see this paper).

As for the second question, if there is such a construction, it should be covered in one of Fremlin's Measure theory volumes.

As for the first question the answer is no.

When $p=1$ we may use the following reasoning. If $\mu$ is $\sigma$-finite, then the space $L_1(\mu)$ is weakly compactly generated. Indeed, by a simple change of measure you may assume that $\mu$ is finite and then the inclusion $L_2(\mu)\to L_1(\mu)$ has dense range. On the other hand, it is well-known that $\ell_1(\Gamma)$ does not embed into a weakly compactly generated space for any uncountable set $\Gamma$, however by the principle of local reflexivity you will find it in some ultrapower of $L_1[0,1]$.

When $p\in (1,2)$, then $\ell_p(\omega_1)$ does not embed into $L_p(\mu)$ with $\mu$ finite

P. Enflo, H. P. Rosenthal, Some results concerning $L_p(\mu)$-spaces. J. Functional Analysis 14 (1973), 325–348.

This was extended to $p\in (2,\infty)$ by Bill Johnson and Gideon Schechtman (see this paper).

As for the second question, if there is such a construction, it should be covered in one of Fremlin's Measure theory volumes.

As for the first question the answer is no.

When $p=1$ we may use the following reasoning. If $\mu$ is $\sigma$-finite, then the space $L_1(\mu)$ is weakly compactly generated. Indeed, by a simple change of measure you may assume that $\mu$ is finite and then the inclusion $L_2(\mu)\to L_1(\mu)$ has dense range. On the other hand, it is well-known that $\ell_1(\Gamma)$ does not embed into a weakly compactly generated space for any uncountable set $\Gamma$, however by the principle of local reflexivity you will find it in some ultrapower of $L_1[0,1]$.

When $p\in (1,2)$, then $\ell_1(\Gamma)$ does not embed into $L_p(\mu)$ with $\mu$ finite for $\Gamma$ uncountable; this is proved in:

P. Enflo, H. P. Rosenthal, Some results concerning $L_p(\mu)$-spaces. J. Functional Analysis 14 (1973), 325–348.

This was extended to $p\in (2,\infty)$ by Bill Johnson and Gideon Schechtman (see this paper).

As for the second question, if there is such a construction, it should be covered in one of Fremlin's Measure theory volumes.

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Tomasz Kania
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