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2010 If Stockholm Open – Doubles
Tennis tournament
Tennis tournament
AHO Jean-Julien Rojer FIN Jarkko Nieminen
Main article: 2010 If Stockholm Open
Bruno Soares and Kevin Ullyett were the defending champions. Ullyett retired from professional tennis in 2009.
As a result, Soares partnered up with his compatriot Marcelo Melo, but they lost in semifinals against Eric Butorac and Jean-Julien Rojer.
Butorac and Rojer won their second doubles title in this year (last win in Tokyo). They defeated Johan Brunström and Jarkko Nieminen 6–3, 6–4 in the final.
Seeds
- POL Łukasz Kubot / AUT Oliver Marach (first round)
- RSA Wesley Moodie / BEL Dick Norman (semifinals)
- SWE Robert Lindstedt / ROU Horia Tecău (first round)
- SWE Simon Aspelin / AUS Paul Hanley (first round)
Draw
Draw
| RD1-seed01=1 | RD1-team01=POL Ł Kubot AUT O Marach | RD1-score01-1=4 | RD1-score01-2=6 | RD1-score01-3=[13] | RD1-seed02= | RD1-team02=MEX S González GER P Marx | RD1-score02-1=6 | RD1-score02-2=4 | RD1-score02-3=[15]
| RD1-seed03= | RD1-team03=ESP M Granollers ESP F López | RD1-score03-1=6 | RD1-score03-2=4 | RD1-score03-3=[5] | RD1-seed04= | RD1-team04=BRA M Melo BRA B Soares | RD1-score04-1=1 | RD1-score04-2=6 | RD1-score04-3=[10]
| RD1-seed05=4 | RD1-team05=SWE S Aspelin AUS P Hanley | RD1-score05-1=64 | RD1-score05-2=3 | RD1-score05-3= | RD1-seed06= | RD1-team06=USA E Butorac AHO J-J Rojer | RD1-score06-1=77 | RD1-score06-2=6 | RD1-score06-3=
| RD1-seed07= | RD1-team07=FRA A Clément AUT J Knowle | RD1-score07-1=4 | RD1-score07-2=2 | RD1-score07-3= | RD1-seed08=WC | RD1-team08=USA B Battistone SWE A Siljeström | RD1-score08-1=6 | RD1-score08-2=6 | RD1-score08-3=
| RD1-seed09= | RD1-team09=GER C Kas GER M Kohlmann | RD1-score09-1=65 | RD1-score09-2=6 | RD1-score09-3=[16] | RD1-seed10=ALT | RD1-team10=ESP D Gimeno-Traver ITA A Seppi | RD1-score10-1=77 | RD1-score10-2=3 | RD1-score10-3=[14]
| RD1-seed11= | RD1-team11=SWE J Brunström FIN J Nieminen | RD1-score11-1=6 | RD1-score11-2=6 | RD1-score11-3= | RD1-seed12=3 | RD1-team12=SWE R Lindstedt ROU H Tecău | RD1-score12-1=3 | RD1-score12-2=1 | RD1-score12-3=
| RD1-seed13= | RD1-team13=GER B Becker GER M Berrer | RD1-score13-1=4 | RD1-score13-2=7 | RD1-score13-3=[5] | RD1-seed14=WC | RD1-team14=FIN H Kontinen SWE M Ryderstedt | RD1-score14-1=6 | RD1-score14-2=5 | RD1-score14-3=[10]
| RD1-seed15= | RD1-team15=ISR J Erlich AUS J Kerr | RD1-score15-1=7 | RD1-score15-2=65 | RD1-score15-3=[7] | RD1-seed16=2 | RD1-team16=RSA W Moodie BEL D Norman | RD1-score16-1=5 | RD1-score16-2=77 | RD1-score16-3=[10]
| RD2-seed01= | RD2-team01=MEX S González GER P Marx | RD2-score01-1=7 | RD2-score01-2=3 | RD2-score01-3=[5] | RD2-seed02= | RD2-team02=BRA M Melo BRA B Soares | RD2-score02-1=5 | RD2-score02-2=6 | RD2-score02-3=[10]
| RD2-seed03= | RD2-team03=USA E Butorac AHO J-J Rojer | RD2-score03-1=7 | RD2-score03-2=64 | RD2-score03-3=[10] | RD2-seed04=WC | RD2-team04=USA B Battistone SWE A Siljeström | RD2-score04-1=5 | RD2-score04-2=77 | RD2-score04-3=[8]
| RD2-seed05= | RD2-team05=GER C Kas GER M Kohlmann | RD2-score05-1=5 | RD2-score05-2=3 | RD2-score05-3= | RD2-seed06= | RD2-team06=SWE J Brunström FIN J Nieminen | RD2-score06-1=7 | RD2-score06-2=6 | RD2-score06-3=
| RD2-seed07=WC | RD2-team07=FIN H Kontinen SWE M Ryderstedt | RD2-score07-1=66 | RD2-score07-2=2 | RD2-score07-3= | RD2-seed08=2 | RD2-team08=RSA W Moodie BEL D Norman | RD2-score08-1=78 | RD2-score08-2=6 | RD2-score08-3=
| RD3-seed01= | RD3-team01=BRA M Melo BRA B Soares | RD3-score01-1=63 | RD3-score01-2=4 | RD3-score01-3= | RD3-seed02= | RD3-team02=USA E Butorac AHO J-J Rojer | RD3-score02-1=77 | RD3-score02-2=6 | RD3-score02-3=
| RD3-seed03= | RD3-team03=SWE J Brunström FIN J Nieminen | RD3-score03-1=6 | RD3-score03-2=6 | RD3-score03-3= | RD3-seed04=2 | RD3-team04=RSA W Moodie BEL D Norman | RD3-score04-1=2 | RD3-score04-2=3 | RD3-score04-3=
| RD4-seed01= | RD4-team01=USA E Butorac AHO J-J Rojer | RD4-score01-1=6 | RD4-score01-2=6 | RD4-score01-3= | RD4-seed02= | RD4-team02=SWE J Brunström FIN J Nieminen | RD4-score02-1=3 | RD4-score02-2=4 | RD4-score02-3=
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