Lakers forward LeBron James and Celtics guard/forward Jaylen Brown have been named the NBA’s Players of the Week, the league announced on Monday (via Twitter).
James, the Western Conference’s winner, helped guide L.A. to a 2-1 record last week, averaging 30.3 points, 8.7 assists, 4.7 rebounds, 1.3 steals and 1.0 blocks with a .569/.500/.600 shooting line in his three appearances (37.6 MPG). The four-time MVP also eclipsed 40,000 career points in Saturday’s loss to the Nuggets, expanding his lead as the NBA’s all-time leading scorer.
Brown, who won for the East, had a strong week as well, averaging 28.3 PPG, 5.3 RPG, 3.0 APG and 1.0 SPG on .623/.471/.846 shooting in 29.3 MPG in convincing victories over Philadelphia, Dallas, and Golden State. The Celtics, who have won 11 straight games, hold (by far) the best record in the NBA at 48-12, six games ahead of the Thunder.
According to the NBA, the other nominees in the West were Luka Doncic, Shai Gilgeous-Alexander, Brandon Ingram, Nikola Jokic and Victor Wembanyama, while Bam Adebayo, Jimmy Butler, Giannis Antetokounmpo, Paolo Banchero, Tyrese Maxey, Evan Mobley, and Brown’s teammate Jayson Tatum were nominated in the East (Twitter links).