Suppose that we consider instead the language

$$ L = { langle M rangle : text{$M$ does not accept $langle M rangle$ in space $f(langle M rangle)$} }. $$

We want to show that $L notin mathsf{SPACE}(o(f(n))$, that is, that if $M$ uses space $o(f(n))$ then $L(M) neq L$. This should be the case since $$langle M rangle in L Leftrightarrow langle M rangle notin L(M).$$

But is this really true? According to the definition of $L$, $langle M rangle in L$ iff $M$ does not accept $langle M rangle$ *in space $f(langle M rangle)$*. It could be that $langle M rangle in L$ and $M$ accepts $langle M rangle$ *using more than $f(langle M rangle)$ space*. The latter could actually happen, since we are only guaranteed that $M$ uses space $g(n)$ for some function $g(n) = o(f(n))$, which does not preclude $g(|langle M rangle|) > f(|langle M rangle|)$ at the *particular* value $|langle M rangle|$.

Adding the padding fixes this issue: it cannot be that $g(|(langle M rangle, 10^k)|) > f(|(langle M rangle, 10^k)|)$ for all $k$, since this would contradict $g(n) = o(f(n))$.