Is the sequence equibounded on $[0,1]$? I've got the following question to pose... it seems to me reasonable for the answer to be positive, anyway I'd like a neat argument to prove it... the question is this.. suppose we are given a sequence $f_n$ of real continuous function defined on $[0,1]$, and we know that $$\lim_{n\to+\infty}\int_{[0,1]}f_n\mathrm d\mu=0,$$ for any Borel measure which is positive and bounded on $[0,1]$. Then is it true that the sequence $f_n$ is equibounded?
Thanks in advance for your replies..
 A: 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 $n_1<n_2<\cdots<n_{k-1}$ 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 $\max\limits_{x\in[0,1]}|f_n(x)|$.)
A: Here is an argument using way more abstract machinery than Jonas Meyer's.
First, note by the Hahn-Jordan decomposition theorem that any finite signed measure $\mu$ on $[0,1]$ can be written $\mu = \mu^+ - \mu^-$ for positive measures $\mu^+$,$\mu^-$.  Thus in fact we have $\int f_n \,d\mu \to 0$ for every finite signed measure $\mu$.
The Riesz representation theorem states that the dual of the Banach space $C([0,1])$ is precisely the space of finite signed measures.  So in fact our condition is that $f_n \to 0$ weakly in $C([0,1])$.
Now it follows from the uniform boundedness principle that a weakly convergent sequence in a normed space is bounded in norm.  Thus we have $\sup_n \|f_n\|_\infty < \infty$ which is to say that the $f_n$ are equibounded.
(Sketch of the last step: for each $n$, the map $\ell_n(\mu) = \int f_n\,d\mu$ is a continuous linear functional on $C([0,1])^*$.  By assumption, we have for each $\mu$ that $\sup_n |\ell_n(\mu)| = \sup_n \left|\int f_n\,d\mu\right| < \infty$ since $\{\int f_n\,d\mu\}$ is a convergent sequence of real numbers and hence bounded.  So by the uniform boundedness principle, $\sup_n \|\ell_n\|_{C([0,1])^{**}} < \infty$.  But by the Hahn-Banach theorem, we have $\|\ell_n\|_{C([0,1])^{**}} = \|f_n\|_\infty$, i.e. the natural map $C([0,1]) \to C([0,1])^{**}$ is an isometry.)
