Report: Floyd Mayweather to fight Andre Berto in September on CBS

Undefeated welterweight world champion Floyd Mayweather reportedly to fight fellow welterweight Andre Berto on Sept. 12 on CBS

Undefeated welterweight world champion Floyd Mayweather is set to fight fellow welterweight Andre Berto on Sept. 12, reports Michael Woods of the Sweet Science. No confirmation has been made from the Mayweather or Berto camps.

According to the report, the fight will be televised on CBS, with Mayweather (48-0, 26 KOs) slated to take home his usual purse, which will be around $30-35 million.

The fight will be marketed as a “give back” to fans who were less than thrilled over paying upwards of $99 to see Mayweather’s fight against Manny Pacquiao via pay-per-view.

Mayweather blamed the boring fight on Pacquiao and his inability to land punches. Pacquiao says he was injured during training in the weeks before the welterweight unification bout and had surgery to repair a torn right rotator cuff after the fight.

Welterweight contender Amir Khan (30-3, 19 KOs) had expressed interest in a fight and IBO/WBA middleweight champion Gennady Golovkin (33-0, 30 KOs) recently said that he would drop down to 154 pounds to fight the 11-time world champion.

Mayweather’s reported bout against Berto (30-3, 23 KOs) is the last in his lucrative six-fight deal with Showtime/CBS that he signed in 2013. He has won each of previous five bouts by decision.

Mayweather last fought in May and earned a 12-round unanimous decision over Pacquiao, retaining the WBA and super welterweight WBC titles and winning the WBO championship from Pacquiao.

The fight broke pay-per-view and live-gate records and generated more than $600 million in revenue, but Mayweather was stripped of his World Boxing Organization belt after missing the deadline to pay the sanctioning body a $200,000 fee.

The 31-year-old Berto last fought in March, scoring a sixth-round TKO over Josesito Lopez, winning the interim WBA Welterweight title. Berto is just 3-3 in his last six fights.

- Scooby Axson