ESPN’s Adrian Wojnarowski reports that free agent forward Paul George has “committed to sign a deal” to remain with the Oklahoma City Thunder.

Update: it’s a four-year, year, $137 million max deal:

And if you don’t believe Woj, how about George himself?

George was at a “summer hype house party” hosted by Thunder star Russell Westbrook on Saturday night, and told everyone in attenndance, “I’m here to stay.”

So there you go. It’s been long assumed that George would end up with the Los Angeles Lakers, but in the end he’ll stay with the team he was traded to last offseason by the Indiana Pacers. The Lakers’ pursuit of LeBron James and Kawhi Leonard could’ve also caused that to fall apart, of course.

With George and Westbrook, the Thunder entered the 2017-18 viewed as a potential threat to the Golden State Warriors and Houston Rockets, but were a huge disappointment and lost in the first-round of the playoffs to the Utah Jazz.

The Thunder will need more than production from Westbrook and George to take the next step in a loaded Western Conference, but having two stars is a pretty big first step to title contention in the NBA. It also has to make the Thunder feel better than they got more than just one season out of George, especially after trading Victor Oladipo for him.

About Matt Clapp

Matt is an editor at The Comeback. He attended Colorado State University, wishes he was Saved by the Bell's Zack Morris, and idolizes Larry David. And loves pizza and dogs because obviously.

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