Нападающий "Барселоны" и сборной Аргентины Лионель Месии стал лучшим футболистом 2019 года по версии Международной федерации футбола (ФИФА). Об этом сообщается в Twitter-аккаунте организации.
В борьбе за награду аргентинец обошел форварда «Ювентуса» Криштиану Роналду и защитника «Ливерпуля» Вирджила ван Дейка. Церемония награждения проходит в Милане в театре «Ла Скала».
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