Misunderstanding of proof with transitivity of logspace reduction I read a commonly known proof of transitivity of logspace reduction:
Lemma 6.: http://www.cs.au.dk/~arnsfelt/CT10/scribenotes/lecture5.pdf
.  In this proof it is used following trick: Let some turing machine $M_f$ for language $f$ compute only $j$-th bit of output.  Ok, $M_f$ is logspace machine, however tape with output can require $O(n)$ space. Hence, we can't get $j$-th  bit in such easy way.   

Where am I wrong ?
$L_f$, by definition, gives you the output of $f(x)$ for a single bit $j$. Note that we do not have to copy $x$, so we're only storing $bin(j)$ and the $j$-th bit of  $f(x)$, which is in logarithmic space. Since $M_f$ computes $L_f$ with logarithmic space, we can simply get the $j$-th bit. This means that $M_h$ can work in logspace in total, as it only asks and stores a single bit of input from $M_f$ every time.