Applications of Dominated/Monotone convergence theorem Consider a measure $\mu$ on $(\mathbb{R}, \mathcal{B}(\mathbb{R}))$ where $\mathcal{B}(\mathbb{R})$ is the Borel $\sigma$-algebra on $\mathbb{R}$. Consider the function $f: [0,\infty)\rightarrow \{1,0\}$, $f(u)=1_{[u=0]}$ where 
$$
1_{[u=0]}=
\begin{cases}
1 \text{ if $u=0$}\\
0 \text{ otherwise}
\end{cases}
$$
Consider the function $g: [0,\infty)\rightarrow [0,\infty)$, $g(u)=u$.
Statement: 
(a) it is possible to construct a sequence of measurable continuous functions $\{f_m(\cdot)\}_m$ such that $\lim_{m \rightarrow \infty}f_m(u)=f(u)$ $\forall u$, and $1\geq f_m(u)\geq f_{m+1}(u)$ $\forall u,m$ (hence, we can apply dominated convergence theorem). 
(b) it is possible to construct a sequence of measurable continuous functions $\{g_m(\cdot)\}_m$ such that $\lim_{m \rightarrow \infty}g_m(u)=g(u) \forall u$ and $0\leq g_m(u)\leq g_{m+1}(u)$ $\forall u,m$ (hence, we can apply monotone convergence theorem).  
(from van der Vaart "Asymptotic Statistics" proof Lemma 6.4 p.89). 
Question: on which result this statement is based? 
My attempt: I know the following two results but I'm not sure whether they fit for the statement above
(1) Consider $h: \mathbb{R}\rightarrow [0,\infty)$. Construct the partition of $[0,\infty)$ in $2^{2m}+1$ intervals of length $2^{-m}$. Let $I_{m.k}$ be the $k$-th interval. Define $h_m(u)=\sum_{k=1}^{2^{2m}+1}\frac{k-1}{2^m}1_{[f^{-1}(I_{m,k})]}$. Then $\lim_{m \rightarrow \infty}h_m(u)=h(u)$ $\forall u$. 
(2) Consider an open set $G \in \mathcal{B}(\mathbb{R})$. Then, there exists a sequence of functions $h_m(\cdot)$ such that $0\leq h_m(u)\leq h_{m+1}(u)$ $\forall m,u$, $\lim_{m \rightarrow \infty}h_m(u)=1_{[u \in G]}$ $\forall u$ (implying that $h_m(u) \leq 1$ $\forall m,u$). 
 A: Sorry for the long post:
As stated, neither of these results are useful to prove the statement, as in both results the functions $h_m$ are not guaranteed to be continuous. In the first result, the functions $h_m$ defined there are not continuous in either case. For statement a), we have
$$
f^{-1}(I_{m,k})=
\begin{cases}
\Bbb R\setminus\{0\} & k=1\\
\{0\} & k=2^m+1\\
\phi & ~
\end{cases}
$$
so $h_m(u)=1_{[u=0]}$, which is not continuous at 1. For statement b) we have $g^{-1}(I_{m,k})=I_{m,k}$ so $h_m(u)=\sum_{k=1}^{2^{2m}+1}\frac{k-1}{2^m}1_{I_{m,k}}$ which is not continuous in the extremes of intervals $I_{m,k}$. 
The second result could help proving part a) of the statement if functions $h_m$ were known to be continuous: we can 
consider $G=(0,+\infty)$. Then we have a sequence $(h_m)$ s.t: for all $u$ it is increasing, $0\leq h_m(u)\leq 1$ and $\lim_{m\to \infty}h_m(u)=1_{[u>0]}$. Then the sequence $(f_m)$ given by $f_m=1-h_m$ verifies almost all the required conditions, save for the continuity. 

I should mention we can prove the statement without involving other results. For a), we can consider the sequence $(f_m)$ s.t. for all $m$, 
$$
f_m(u)=
\begin{cases}
0 & u>1/m\\
1-u/m & 0\leq u \leq 1/m
\end{cases}
$$ 
We can check that: for all $m, f_m$ is continuous (so measurable) and, for all $u$, $(f_m(u))_m$ is decreasing, and $\lim_{m\to \infty} f_m(u)=f(u)$.
For b), we can take the sequence $(g_m)$ s.t. for all $m, g_m=g$. By the way, do you mean to approach $g$ by a sequence of continuous functions? Or by a sequence of integrable continuous functions? In that case, we can take the sequence $(g_m)$ s.t. for all $m$, 
$$
g_m(u)=
\begin{cases}
0 & u>2m\\
u & 0\leq u \leq m\\
m-u & m\leq u \leq 2m
\end{cases}
$$ 
We can check: for all $u$, $(g_m(u))_m$ is increasing and $\lim_{m\to\infty}g_m(u)=g(u)$, and for all $m, g_m$ are continuous and integrable, with $\int g_md\mu\leq m\mu([0,2m])$. 
A: For (a) the sequence $f_m(u) = 1/(1+u)^m$ will work.
