shell script search for one or more whitespace regular expression

I have a shell script where I run du -sh on a directory, and if I want to find all Mb files, I run this line in my script:

      du -sh $DIR/* 2>/dev/null|grep -v [0-9][Mm]|sort -nr
The problem I have with this line is that if there's files with other digits and M in the name, it matches those too, like this:

146M    /tmp/file4ZURF4
117M    /tmp/fileoVe4mF
85M     /tmp/filezN0eoA
84K     /tmp/fileT8mZQr.csv
84K     /tmp/file4MbmT3.csv

So I want to match the whitespace that occurs after the M in 146M.  But I can't find the exact regular expression to do this in a shell script.  Can someone pass this along?  Thanks!
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try this
      du -sh $DIR/* 2>/dev/null|grep -v [0-9][Mm][\s]|sort -nr
Combination '<digit>M<space>' may be met in the name of file, too...

Need to find all strings, which begin from digits. It is symbol '^' in regexp.
Also, there may be more then one digit and decimal point. We need to use symbol '*'.
Besides that, option '-v' (or --invert-match) inverts the sense of matching, to select non-matching lines (according to man grep).
So, the final expression:

du -sh $DIR/* | grep ^[.0-9]*[Mm] | sort -nr

'.' in expression '[.0-9]' means decimal point, may vary in different locale.
In GNU regex it should be

 du -sh $DIR/* 2>/dev/null | grep "[0-9][Mm]\>" | sort -nr

- Why "grep -v" ? Doesn't make sense to me, unless yo're trying to exclude all MB files

- What if a filename ends in e.g. ...9M ?

In this case I'd recommend

du -sh $DIR/* 2>/dev/null  | grep -E  "^[0-9.]+[Mm]\>" |sort -nr



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