Let $(\Omega,\Sigma,\mu)$ be a sample space and let $L^2= \lbrace f:\Omega \rightarrow R / \int f^2d\mu <\infty \rbrace$ be a Hilbert space. Let $L_n=L^2\times L^2 \times .... \times L^2$ ($n$ times) be a vector space with inner product $\langle X,Y \rangle = \sum_{i=1}^{n} \int X_iY_id\mu$ where $X=(X_1,...,X_n)$ and $Y=(Y_1,...,Y_n)$.
Also consider:
$X \geq 0$ if and only if $X_i \geq 0$, $\forall \: i \in \lbrace 1,..,n \rbrace$.
$X = Y$ if and only if $X_i = Y_i$ almost everywhere, $\forall \: i \in \lbrace 1,..,n \rbrace$.
If $F:L_n \rightarrow R$ is a linear functional and it's positive (i.e. $F(X)\geq0$ for $X\geq0$) then $F$ is continuous. (i.e. $\Vert X_n - X \Vert \rightarrow 0$ imply $F(X_n) \rightarrow F(X)$)
The following works in the case $n=1$. The general case easily follows.
Assume that $F$ is not continuous. Then for every $k \in \Bbb{N}$, there is some $f_k \in L^2$ with $\Vert f_k \Vert_2 \leq \frac{1}{2^k}$ and $F(f_k) \geq 2^k$ (show this!).
Using e.g. monotone convergence, one can show (do it!) $g := \sum_{\ell = 1}^{\infty} |f_\ell| \in L^2$ with $\Vert g \Vert_2 \leq 1$.
But now $0 \leq g \pm f_k$ for every $k$ and thus $0 \leq F(g \pm f_k)$ for all $k$.
Why does this yield a contradiction?