The following fact is known:



> 
> If there is a measurable cardinal, then there are only countably many constructible reals.
> 
> 
> 


It is also known that if $ZFC$ + "There is a (two-valued) mesurable cardinal" is consistent, then $ZFC$ + "There exists a (two-valued) measurable cardinal" + $CH$ is also consistent.


If one replaces "There exists a (two-valued) measurable cardinal" with "The cardinality of the continuum is a real-valued measurable cardinal", is 


i)$ZFC$ + "The cardinality of the continuum is a real-valued measurable cardinal" + $CH$ is consistent if $ZFC$ + "The cardinality of the continuum is a real-valued measurable cardinal" is consistent (it is not, by a result of Banach and Kuratowski)? Also,


ii) Does $ZFC$ + "The cardinality of the continuum is a real-valued measurable cardinal" imply that there are only countably many constructible reals?


Since i) is inconsistent, how is it possible that $ZFC$ + "The cardinality of the continuum is a real-valued measurable" implies $\lnot$$CH$ while $ZFC$+ "There exists a (two-valued) measurable cardinal" is consistent with $CH$ or $\lnot$$CH$?



Real-valued measurability vs. Two-valued measurability in determining whether $CH$ holds or not
First of all, it is **not** the case that if ZFC is consistent, then ZFC + CH + "There is a measurable" is consistent; a measurable cardinal has *much* greater consistency strength than ZFC alone. What **is** true is that if ZFC + "There is a measurable" is consistent, then so is ZFC + CH + "There is a measurable;" I think this is what you mean.


*A bit of clarification, just to pin things down: when we say "If $A$ is consistent, then $B$ is consistent," this is usually shorthand for "$T$ proves $Con(A)\implies Con(B)$" where $T$ is some reasonable weak theory (PRA is almost always enough); such an interpretation is needed to avoid stupidity (if ZFC is consistent, then the sentence "PA is consistent implies ZFC is consistent" is technically true, even though what we **mean** by it clearly isn't). So, elaborating a bit on the previous paragraph, here are a couple things PRA proves:*


* *"If ZFC + "There is a measurable" is consistent, then ZFC + CH + "There is a measurable" is consistent."*
* *"If PRA proves "If ZFC is consistent, then ZFC + "There is a measurable cardinal" is consistent," then ZFC is inconsistent."*


*No, that second one wasn't a typo.*




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Now re: question 1, you ask:



> 
> how is it possible that ZFC + "The cardinality of the continuum is a real-valued measurable" implies $\neg$CH while ZFC+ "There exists a (two-valued) measurable cardinal" is consistent with CH since the two theories are equiconsistent?
> 
> 
> 


There's no tension here - equiconsistent theories can disagree. Equiconsistency doesn't mean that they're consistent *together*, but rather that each is consistent if and only if the other is consistent. For example, ZFC+CH and ZFC+$\neg$CH are equiconsistent, but clearly inconsistent with each other.


So there's no issue.




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Re: question 2, off the top of my head I'm not sure, but if memory serves Solovay showed that if there is a real-valued measurable, then there is an inner model with a measurable *(it is certainly true that he showed that "there is a real-valued measurable" and "there is a measurable" are equiconsistent over ZFC; but it's possible he build an inner model of a forcing extension, instead of an actual inner model, which would be a problem here)*. If my memory is correct, then the answer to 2 is "yes:" letting $M$ be that inner model, we have $M\models \vert\mathbb{R}^L\vert=\aleph\_0$. But $L$ (hence $\mathbb{R}^L$) and $\aleph\_0$ are absolute, so $V\models\vert\mathbb{R}^L\vert=\aleph\_0$.


*Note that this is stronger than what you ask for in (2): the conclusion is that if there is a real-valued measurable **at all**, regardless of its comparison with the continuum, then $\mathbb{R}^L$ is countable.*


