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I read in a book that if $X_n \geq 0$, and $X_n \to X$ with probability $1$, then if $E(X_n) \leq 1$, by Fatou's Lemma, $E(X)$ exists and $E(X) \leq 1$.

I am wondering how to explicitly calculate this. Would it be:

Since $X_n \to X$ almost surely, then $\liminf_{n \to \infty}X_n = X$ as a corollary. Then:

\begin{align} E(X) &= E\left(\liminf_{n \to \infty}X_n\right) \ \ \text{(by almost sure convergence)}\\ &\leq \liminf_{n \to \infty}E(X_n) \ \ \text{(by Fatou's Lemma)}\\ &\leq \liminf_{n \to \infty} 1 \ \ \text{(Given sequence bound)} \\ &= 1 \ \ \text{(trivial result)} \end{align}

Furthermore, the book states that IF we tried to do this by the Dominated Convergence Theorem to show $E(X)$ exists, then we need to show:

$$ Y_n = \sup_n E(X_n) $$

has finite expectation and that such a bound requires joint distribution information of the $X_n$ in contrast to only the marginal distributions of $X_n$ by using Fatou's Lemma, why is that?

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    Before "Furthermore" you are spot on. After it, please replace $Y_n=\sup E(X_n)$ by $Y=\sup X_n$ (two changes), and proceed. – 2017-01-12
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    @Did I see, so basically the difference is the DCT requires the supremum of the expectation of $X_n$ while for Fatou we are concerned about the expectation of the supremum of $X_n$? – 2017-01-12

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