Lebesgue integral - convergence of supremums Let $\newcommand{\d}{\mathrm{d}}(f_{a, n})_{a \in (0, 1), \, n \in \mathbb{N}} \subseteq L^1$ and $f_{a} \in L^1$ for every $a\in (0,1)$. Suppose that for each parameter $a$ one has that
$$
f_{a,  n} \stackrel{n\to\infty}{\longrightarrow} f_a.$$ 
Assume further that  there exist constants $C_1, C_2$ such that for all $a \in (0, 1), \, n \in \mathbb{N}$
$$
C_1 \leq \int f_{a,  n} \,\d\mu\leq C_2.
$$
Is it then true that
$$
\sup_{a \in (0, 1)} \int f_{a,  n}\,\d\mu \stackrel{n\to\infty}{\longrightarrow} \sup_{a \in (0, 1)} \int f_a \,\d\mu ?
$$

Let us consider $L^1(0,1)$ with the Lebesgue measure $\newcommand{\d}{\mathrm{d}}\lambda$. Then we can define 
\begin{align}
f_{a,n} := \begin{cases}
0, & \text{if}\; a \notin \{2^{-i}: i \in \mathbb{N}\}, \\
1, & \text{if there exists an}\; i \in \mathbb{N} \;\text{sucht that}\; a=2^{-i} \;\text{and}\; n\leq i.
\end{cases}
\end{align}
For each $a\in (0,1)$ we have that $\{f_{a,n}\}_{n\in\mathbb{N}}$ converges in $L^1$-norm to the zero-function. This means that 
$$\sup_{a\in (0,1)}\int f_a \,\d \lambda=0.$$
On the other hand if we fix $n\in \mathbb{N}$, we obtain: 
$$\sup_{a\in (0,1)}\int f_{a,n}\,\d\lambda=1.$$
