When we attempt a transformation to a new variable \( x\rightarrow y \) we have to conserve the probability $$ \begin{equation*} p(y)dy=p(x)dx, \end{equation*} $$ which for the uniform distribution implies $$ \begin{equation*} p(y)dy=dx. \end{equation*} $$ Let us assume that \( p(y) \) is a PDF different from the uniform PDF \( p(x)=1 \) with \( x \in [0,1] \). If we integrate the last expression we arrive at $$ \begin{equation*} x(y)=\int_0^y p(y')dy', \end{equation*} $$ which is nothing but the cumulative distribution of \( p(y) \), i.e., $$ \begin{equation*} x(y)=P(y)=\int_0^y p(y')dy'. \end{equation*} $$