Limit integral switch for nondecreasing functions I have a sequence of functions $f_n:[a,b] \to \mathbb{R}$ converging pointwise to a continuous function $f$.
If every $f_n$ is nondecreasing [$f_n(x) \leq f_n(y)$ for $x < y$] then can it be shown that
$$\lim_{n \to \infty}\int_a^b f_n = \int_a^b f$$
I thought of monotone convergence or dominated convergence theorems but I don't have $f_n$ increasing with $n$ and I can't be sure $|f_n(x)| < |f(x)|$. In fact I don't even know where $f$ may be positive or negative.

We can show, in this case, that the convergence is uniform and switching the limit and integral is permissible.
Since $f$ is uniformly continuous on $[a,b]$, for any $\epsilon > 0$  there is a partition where on each subinterval we have 
$$\tag{*}-\epsilon/2 < f(x_j) - f(x_{j-1}) < \epsilon/2$$ 
Any $x \in [a,b]$ belongs to some subinterval $[x_{j-1},x_j]$. By pointwise convergence and monotonicity of $f$ (since $f_n \to f$ and $f_n$ is nondecreasing), there exists $N$ depending on $x_1,\ldots,x_j  \ldots$ and $\epsilon$ but not $x$ such that for $n > N$
$$f(x_{j-1}) - f(x_j) - \epsilon/2 < f_n(x_{j-1}) - f(x_j) \\< f_n(x) - f(x) \\< f_n(x_j) - f(x_{j-1}) < f(x_j) - f(x_{j-1}) + \epsilon/2.$$
Using (*) it follows that $|f_n(x) - f(x)| < \epsilon$ for all $x \in [a,b],$ and convergence is uniform.
