When I initially read about this, I thought it must be due to the cardinality of a measurable cardinal being "too high to fit in L". But this is obviously false, because the measurable cardinal exists as an ordinal in L.
So my question then is "Why isn't this ordinal in L recognized as a measurable cardinal ?".
Or equivalently, "What prevents me from constructing a $\kappa$-complete ultrafilter on this ordinal ?"
Request: Please don't repeat the formal proof of the theorem as this is available in standard textbooks.
What I am looking for is something more intuitive, similar to:
Q:"Why are the real numbers uncountable ?"
A:"Because every time you try to construct a bijection from naturals to reals, you can always find a real number which is not in the image of the bijection"
Here is the intuition in a nutshell.
If you have a measure, you can take an ultrapower of the universe to define a proper submodel. Why proper? We can show that a measure is never itself an element of the ultrapower it generates. But this would be an inner model smaller than $L$. This is impossible, in part because the ultrapower satisfies $V=L$, and therefore must be $L$.
So the measurable cardinal is itself in $L$, and sure it's inaccessible and even more, when considered in $L$. But its measure(s) are not in $L$. So it is not measurable in $L$.