General Fatou's Lemma How can I proof the general Fatou's Lemma without using the Monotone convergence Theorem.   
Lemma:
Let $(X,\mathcal{M},\mu)$ be a measure space and $\{f_n\}$ a non-negative measurable sequence. Then  
$$ \int_X \liminf_{n\to\infty}f_n~d\mu \leq \liminf_{n\to\infty}\int_X f_n~d\mu.$$

Here is one proof based on the bounded convergence theorem, adapted from Durrett.
Define $g_n(x) = \inf_{m\geq n} f_m(x)$. So, $f_n \geq g_n$ and $g_n \uparrow g(x) := \liminf_n f_n(x)$ as $n \to \infty$.
By monotonicity of the integral, we know that $\newcommand{\du}{\,\mathrm d \mu} \int f_n \du \geq \int g_n \du$, whence 
$$\liminf_n \int f_n \du \geq \liminf_n \int g_n \du \>.$$
Suppose $X_n \uparrow X$ where $\mu(X_n) < \infty$. By the bounded convergence theorem, for fixed $m$, we have
$$
\liminf_n \int g_n \du \geq \int_{X_m} g_n \wedge m \du \to \int_{X_m} g \wedge m \du \>,
$$
since the integrand in the middle is bounded and converges to the integrand on the right.
But, then
$$
\liminf_n \int g_n \du \geq \sup_m \int_{X_m} g \wedge m \du = \int \liminf_n f_n \du \>.
$$
Since $\liminf_n \int f_n \du \geq \liminf_n \int g_n \du$, we are done.
