Here is a proof by contradiction. Suppose that the condition holds and $(f_n)$ is not equibounded. By considering the Dirac measures $\delta_x$ for each $x\in[0,1]$, we see that $f_n\to 0$ pointwise. Let $n_1\in\mathbb N$ and $x_1\in[0,1]$ be such that $|f_{n_1}(x_1)|=\max\limits_{x\in[0,1]}|f_{n_1}(x)|\geq 4$. There is an $n_2>n_1$ and $x_2\in[0,1]$ such that:
- for all $n\geq n_2$, $|f_n(x_1)|<1$;
- $|f_{n_2}(x_2)|=\max\limits_{x\in[0,1]}|f_{n_2}(x)|\geq 4^2$.
Given $n_1 in $\mathbb N$ and $x_1,x_2,\ldots,x_{k-1}\in[0,1]$ such that:
- for each $j\in\{2,\ldots,k-1\}$, for all $n\geq n_j$, $|f_n(x_{j-1})|<1$;
- for each $j\in\{1,\ldots,k-1\}$, $|f_{n_j}(x_j)|=\max\limits_{x\in[0,1]}|f_{n_j}(x)|\geq 4^j$,
there exists $n_k>n_{k-1}$ and $x_k\in[0,1]$ such that:
- for all $n\geq n_k$, $|f_n(x_{k-1})|<1$;
- $|f_{n_k}(x_k)|=\max\limits_{x\in[0,1]}|f_{n_k}(x)|\geq 4^k$.
Consider the sequences $(f_{n_k})$ and $(x_k)$ to have been defined in this way, and let $m_k=|f_{n_k}(x_k)|=\max\limits_{x\in[0,1]}|f_{n_k}(x)|$.
Let $\mu =\sum\limits_{m=1}^\infty3^{-m}\delta_{x_m}$. Then for each $k\in\mathbb N$,
$\begin{align*} \left|\int f_{n_k}d\mu\right|&=\left|\sum\limits_{m=1}^\infty 3^{-m}f_{n_k}(x_m)\right|\\ &\geq 3^{-k}|f_{n_k}(x_k)|-\sum_{m=1}^{k-1}3^{-m}|f_{n_k}(x_m)|-\sum_{m=k+1}^{\infty}3^{-m}|f_{n_k}(x_m)|\\ &\geq m_k3^{-k}-\sum_{m=1}^{k-1}3^{-m}-m_k\cdot\sum_{m=k+1}^{\infty}3^{-m}\\ &\geq m_k3^{-k}-\frac{1}{2}-m_k\frac{1}{2}3^{-k}\\ &=\frac{1}{2}m_k3^{-k}-\frac{1}{2}\\ &\geq\frac{1}{2}\left(\frac{4}{3}\right)^k-\frac{1}{2}. \end{align*}$
So $\left|\int f_{n_k}d\mu\right|\to+\infty$, and this contradicts $f_nd\mu\to 0$.
(Continuity was used for the existence of $\max\limits_{x\in[0,1]}|f_n(x)|$.)