Manchester United News: Louis Van Gaal Claims Club Should Be 'Happy' With Europa League

Manchester United manager Louis Van Gaal, left, faces Jurgen Klopp's Liverpool on Thursday evening.
Louis van Gaal, left, at Old Trafford, May 2, 2015. The Manchester United manager faces Liverpool manager Jurgen Klopp in the Europa League on Thursday evening. Stephen Pond/Getty Images

Manchester United manager Louis Van Gaal has invited criticism from fans by suggesting that the club should be "happy" to be playing in the UEFA Europa League.

United faces Liverpool on Thursday evening, the first time the two giants of English football have met in Europe, with Van Gaal urging United to stop living in the past and accept its current status.

"Other teams are playing in the Champions League. You are talking about Manchester United and Liverpool and I can say we are happy to play in the Europa League," Van Gaal said.

"It is not normal that one team dominates for 20 years in a row as the champion.

"You have noticed that against FC Midtjylland there were 58,000 [fans], so the fans of Manchester United are appreciating [the Europa League]. I think against Liverpool there will be 75,000 [at Old Trafford). Liverpool also is sold out. It is very important for both teams.

"You have to live in the present, I think."

Van Gaal's United team was knocked out of the UEFA Champions League when it was defeated by Wolfsburg, of Germany, in December, and is struggling to qualify for Europe's elite club competition for next season through the English Premier League.

The Europa League, often criticized for its convoluted format, now, however, carries the tantalizing prize of a place in the Champions League for the winner. Liverpool, in a worse position than United in the Premier League, could benefit equally from a back-door route into Champions League riches for next season.

Though it only won the Champions League twice under Alex Ferguson, in 1999 and 2008, United was a perennial contender in the competition during the fearsome Scot's reign.