Well, the current value of $/ has nothing to do with whether or not "." in a regex will match "\n" in a given string value; only the "s" modifier on the regex will allow "." to match "\n", no matter what $/ is. Try this out:

#!/usr/bin/perl

$/ = undef;
$_ = <DATA>;

while ( /begin(.*?)end/g ) {
   print "\n--- found without 's': ---\n$1\n";
}
while ( /begin(.*?)end/gs ) {
   print "\n--- found using 's': ---\n$1\n";
}

__DATA__
blah
begin 
foo
bar
baz
end
begin foo2
bar2
baz2
end
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