Fatou's lemma and measurable sets

I don't know how can I imply Fatou's lemma for any measurable sets $A_k$, that is, $$\lambda(\liminf A_k)\le\liminf\lambda(A_k).$$ How can I prove it? And is there any example in $\mathbb R$ of sequence of measurable sets $A_k$ such that $A_k\subset[0,1]$, $\lim\lambda(A_k)=1$, but $\liminf A_k=\varnothing$ ?

Thanks for your help!

First, since by monotonicity we have $\lambda(\bigcap_{n=k}^{\infty}A_{n})\leq \lambda(A_{j})$ for all $j\geq k$, it follows that $\lambda(\bigcap_{n=k}^{\infty}A_{n})\leq \inf_{n\geq k}\lambda(A_{n})$.

Using convergence of measure to the nondecreasing sequence of sets $((\bigcap_{n=k}^{\infty}A_{n}))_{k=1}^{\infty}$, we obtain:

\begin{align*} \lambda(\liminf A_{n})=\lambda(\bigcup_{k=1}^{\infty}(\bigcap_{n=k}^{\infty}A_{n}))=\lim_{k\to\infty}\lambda(\bigcap_{n=k}^{\infty}A_{n})\leq \lim_{k\to\infty}\inf_{n\geq k}\lambda(A_{n})=\liminf \lambda(A_{n}). \end{align*}

Which is the result that we wanted.