You probably just want a word boundary assertion, \b 
$string =~ s/\b($greekReplace)\b/\\$1 /isg;

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Note that \b doesn't think 3mu has a word boundary between the 3 and mu (because both match \w). If that's a problem, you need to work with the look-around assertions described in perlre.

Update: I should have read the question more carefully, it seems a negative look-behing would be more appropriate: 

s/(?<!a)($greekReplace)/\\$1 /isg;

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In reply to Re: Regex question by moritz
in thread Regex question by memnoch

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