a.s. convergence to constant and expectation

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Is it true that, if $X_n$ converges a.s. to a constant $c$, then $\displaystyle\lim_{n\to\infty}\mathbb{E}(X_n) = c$? If yes, please, can you prove that?

Update: If it is false, is there a condition I can set to prove that it is true?

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Suppose this was true. Then we would get the statement that if $|X_n - X| \to 0$ a.s., then $E(|X_n - X|) \to 0$. This is saying that almost sure convergence implies $L^1$ convergence. But this is false; convergence a.s. does not imply convergence in $L^1$. One counterexample is $X_n = n1_{(0, 1/n)}$ defined on $[0, 1]$. $X_n \to 0$ a.s. but $E(X_n) = \int_{0}^{1}n1_{(0, 1/n)} = 1$.