Approximating the Brownian excursion area Suppose that $(F_n(t))_{t\in[0,1]}$ is a sequence of random continuous real-valued functions such that 
$$(F_n(t))_{t\in[0,1]}\stackrel{d}{\longrightarrow}(e(t))_{t\in[0,1]},$$ 
where $e(t)$ denotes the Brownian excursion.
Is it true that 
$$\mathbb{E}\left[\int_0^1F_n(t)dt\right]\longrightarrow\mathbb{E}\left[\int_0^1e(t)dt\right]?$$
I would like to say yes, since the area function is continuous (but not bounded) and the area of the Brownian excursion is bounded with exponentially high probability. Any help to formalize this intuition or disproving the statement?
 A: The result is false. The problem is, as you pointed, the boundedness of functional $f \to \int f$ which cannot guarantee DCT to hold.
We will use the fact that $\mathbb E[\int_0^1 e(t) dt]$ is finite, where $(e_t)_{t \in [0,1]}$ is B.Excursion
There are two kind of problems.

*

*Firstly, we can have $\mathbb E[\int_0^1 F_n(t) dt] = \infty$ for every $F_n$.

Indeed, for example take $\Omega = (0,1)$ and define $F_n(t)(\omega) = e(t)(\omega) + \frac{1}{\omega n}$. It is easy to see, that for fixed $\omega \in (0,1)$ we have $||F_n(\omega) - e(\omega)||_{\sup} = \frac{1}{\omega n} \to 0$, so that $F_n \to e$ converge almost surely (in $C[0,1]$ space), hence converge in distribution, too. But $$ \mathbb E[\int_0^1 F_n(t) ]= \mathbb E[\int_0^1 e(t)dt] + \int_0^1 \frac{1}{\omega n} d\omega = \infty \not \to \mathbb E[\int_0^1 e(t)dt] < \infty$$


*Even when every $ \int_0^1 F_n(t)dt$ is integrable on $L_1(\Omega)$, the problem can still arise. Take any sequence $X_n:\Omega \to \mathbb R$ such that $X_n \to 0$ almost surely, but $X_n \not \to 0$ in $L_1(\Omega)$. For example $X_n(\omega) = n$ for $\omega \in (0,\frac{1}{n})$ and $X_n(\omega) = 0$ otherwise (again work on $\Omega = (0,1)$. Take any sequence $(c_n)$ of real numbers. Let $F_n(t)(\omega) = e(t)(\omega) + c_nX_n(\omega)$. Similarly as before, for fixed $\omega$ we have $||F_n(\omega) - e(\omega)||_{\sup} = c_n X_n(\omega) \to 0$. But $$ \mathbb E[\int_0^1 F_n(t)dt] = \mathbb E[\int_0^1 e(t)dt] + c_n\mathbb E[X_n] = \mathbb E[\int_0^1 e(t)dt] + c_n$$
As we see, there is still term $c_n$ and we can do nothing to repair it, since $c_n$ can be really arbitrary, even tending to infinity, or alternating, or whatever else.
