This is an important lemma, so I would like to give it a detailed attempt of how such hypothesis is necessary for the lemma to hold. Could you confirm if my understanding is fine?
Let $(X_{n})$ be a sequence of a.s. nonnegative random variables. Then $\mathbb{E}\left(\liminf X_{n}\right) \leq \liminf \mathbb{E}\left(X_{n}\right)$. If there exists a $Y$ such that $X_{n} \leq Y$ a.s. for all $n$ and $\mathbb{E}(Y)<\infty$, then $\mathbb{E}\left(\limsup X_{n}\right) \geq \limsup \mathbb{E}\left(X_{n}\right)$.
Proof:
Lemma 1: $\mathbb E [\inf_{k \ge n} X_k] \le \inf_{k \ge n} \mathbb E [X_k] \le \sup_{k \ge n} \mathbb E [X_k] \le \mathbb E [\sup_{k \ge n} X_k]$ for all $n$.
Proof: We have $\inf_{k \ge n} X_k \le X_t$ and thus $\mathbb E [\inf_{k \ge n} X_k] \le \mathbb E [X_t]$ for all $t\ge n$. Similarly, we obtain $\sup_{k \ge n} X_k \ge X_t$ and thus $\mathbb E [\sup_{k \ge n} X_k] \ge \mathbb E [X_t]$ for all $t\ge n$. The result then follows by taking the $\inf$ and the $\sup$.
Notice that $$\mathbb{E}\left(\liminf X_{n}\right) \leq \liminf \mathbb{E}\left(X_{n}\right) \iff \mathbb{E}\left(\lim_n \inf_{k\ge n} X_{k}\right) \leq \lim_n \inf_{k \ge n} \mathbb{E}\left(X_{k}\right).$$
Let $Y_n = \inf_{k\ge n} X_{k}$. Then $(Y_n)$ is a non-decreasing sequence of non-negative random variable such that $Y_n \nearrow \lim_n \inf_{k\ge n} X_{k}$. By monotone convergence theorem, $\mathbb{E} (Y_n) \nearrow \mathbb{E}\left(\lim_n \inf_{k\ge n} X_{k}\right)$. So it suffices to show that $\mathbb E (Y_n) \le \inf_{k \ge n} \mathbb{E}\left(X_{k}\right)$ for all $n$. However, this follows directly from our Lemma 1.
Lemma 2: Let $( X_{n} )$ be a non-increasing sequence of non-negative random variables such that $X_n \searrow X$ a.s., and $\mathbb E (X_0) <\infty$. Then $\mathbb{E}\left(X_{n}\right) \searrow \mathbb{E}\left( X \right)$.
Proof: We have $(X_0-X_n)$ is a non-decreasing sequence of non-negative random variables such that $(X_0-X_n) \nearrow (X_0-X)$ a.s. By monotone convergence theorem, $\mathbb{E}\left(X_0-X_{n}\right) \nearrow \mathbb{E}\left(X_0- X \right)$. With $\mathbb E (X) \le \cdots \le \mathbb E (X_1) \le \mathbb E (X_0) <\infty$, we have $\mathbb{E} (X_0) - \mathbb{E} (X_{n}) \nearrow \mathbb{E} (X_0) - \mathbb E(X)$. The result then follows.
Notice that $$\mathbb{E}\left(\limsup X_{n}\right) \geq \limsup \mathbb{E}\left(X_{n}\right) \iff \mathbb{E}\left(\lim_n \sup_{k\ge n} X_{k}\right) \geq \lim_n \sup_{k \ge n} \mathbb{E}\left(X_{k}\right).$$
Let $Z_n = \sup_{k\ge n} X_{k}$. Then $(Z_n)$ is a non-increasing sequence of non-negative random variable such that $Z_n \searrow \lim_n \sup_{k\ge n} X_{k}$. Moreover, $\mathbb E (Z_{0}) \leq \mathbb E (Y) < \infty$. By Lemma 2, $\mathbb{E} (Z_n) \searrow \mathbb{E}\left(\lim_n \sup_{k\ge n} X_{k}\right)$. So it suffices to show that $\mathbb E (Z_n) \ge \sup_{k \ge n} \mathbb{E}\left(X_{k}\right)$ for all $n$. However, this follows directly from our Lemma 1.
