There is a theorem of Shelah, stated in the following way:
If all $\Sigma_3^1$ sets of reals are measurable, then $\aleph_1$ is an inaccessible cardinal in $L$.
In some textbooks (for example Schindler, 2014: Set Theory, Exploring Independence and Truth) the theorem is stated slightly different. Instead of inaccessible cardinal in $L$, they say $\aleph_1^V$ is then inaccessible to the reals.
Can someone explain me why this is an equivalent way to state the theorem? Thank you in advance for your help!
The statements are not equivalent. More precisely, $\omega_1$ being inaccessible to the reals is a strictly stronger statement then $\omega_1$ being inaccessible in $L$.
If $\omega_1$ is not inaccessible in $L$, let $\omega_1 = (\alpha^+)^L$ where $\alpha$ is a cardinal in $L$. Then $\alpha$ is countable in $V$ and therefore there exists a real $x \subset \omega$ coding a well order of length $\alpha$ on $\omega$, hence $\omega_1^{L[x]} = \omega_1$.
On the other hand, if $V = L[0^\sharp]$ then $\omega_1$ is inaccessible in $L$.