Example of discontinuous $u(x)$ where $\sum u_{n}(x)\to u(x)$ uniformly in $I \subset \mathbb{R}$ I know that for a series of non-negative, continuous functions $u_{n}(x)$, a sufficient condition for uniform convergence of $\sum u_{n}(x)$ to $u(x)$ is for $u(x)$ to be continuous in $I\subset \mathbb{R}$. 
But I can't think of an example where $\sum f_{n}(x)\to f(x)$ uniformly in $I\subset \mathbb{R}$ but $f(x)$ is discontinuous in $I\subset \mathbb{R}$.
One of $(f_{n}(x))$ must obviously be discontinuous in the interval because (as others have written below) a uniform limit of continuous functions is continuous.
 A: One way to get an example for the clarified question is to let $u_1$ be any discontinuous function on $[0,1]$, like $u_1(x)=0$ if $0\leq x\leq \frac{1}{2}$ and $u_1(x)=1$ if $\frac{1}{2}\lt x\leq 1$, and let $u_n(x)=0$ for $n\geq 2$.  Then the series $\sum u_n(x)$ converges uniformly to $u_1(x)$.
If the $u_n$'s are continuous, you won't be able to find such an example, because a uniform limit of continuous functions is continuous.

Regarding your first statement, it isn't necessarily true if the interval is not compact. For an example of a continuous function that is the nonuniform limit of an increasing sequence of nonnegative continuous functions on an interval, consider $f(x)=1$ on $(0,1)$, and $f_n(x)=1-x^n$.  To phrase this in terms of a series of nonnegative functions, take $u_1=f_1$ and $u_n=f_n-f_{n-1}$ for $n\geq2$.
However, if the interval is compact, then it is true that if a monotonically increasing sequence of continuous functions converges pointwise to a continuous function, then the convergence is uniform.  This is called Dini's theorem.
A: EDIT: If you want discontinuous functions whose sum converges uniformly to a discontinuous function, consider the characteristic function of the rationals: 
$$
   \lambda(x) = \left\{
     \begin{array}{lr}
       1 & : x \in \mathbb{Q}\\
       0 & : x \notin \mathbb{Q}
     \end{array}
   \right.$$
Then let $\lambda_n = \frac{1}{2^{n+1}} \cdot \lambda$. The sum will indeed converge uniformly to $\lambda$ but will clearly be discontinuous. In fact if $\lambda$ is any discontinuous function, this construction will give you your desired result.

However, if the $u_n$'s are all continuous, this cannot happen. Let 
$$g_n = \sum_{k=1}^{n} u_k$$
Then each $g_i$ is countinuous being the sum of continuous functions. But since $g_n \rightarrow u$ uniformly, it must be the case that $u$ is continuous as well. Here's a quick proof:
Suppose we want to show that $u$ is continuous at some $x_0 \in I$. Fix some $\epsilon > 0$. Find $N$ large enough such that $|g_n (x) - u(x) | < \frac{\epsilon}{3}$ for all $x \in I$ and $n \ge N$. Choose some $\delta$ such that $| g_N (x) - g_N (x_0) | < \frac{\epsilon}{3}$ whenever $|x - x_0 | < \delta$, which we can do as $g_N$ is continuous. 
Now we have that:
$$|u(x) - u(x_0)| \le |u(x) - g_N(x)| + |g_N (x) - g_N (x_0) | + |g_N(x_0) - u(x_0)| < \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon $$
by applying the triangular inequality. Hence $u$ is continuous at $x_0$. 
A: Suppose your interval $I = [-1,1]$.
Let $u_n(x) = 2^{-n}$ for $x \in [-1,0)$ and $u_n(x) = 2^{-n} + 1$ for $x \in [0,1]$.  Then $\sum u_n(x) \to u(x)$ uniformly, where $u(x) = 1$ on $[-1,0)$ and $u(x) = 2$ on $[0,1]$.
For general intervals $I$... well, just change the bounds.
A: This follows from the following:
Theorem. If the limit function $f(x)$ of a pointwise convergent sequence of continuous functions $\{f_{n}(x) \}$ is discontinuous, then the convergence of the sequence $\{f_{n}(x) \}$ is nonuniform.
So the contrapositive is: $\{f_{n}(x) \} \to f \ \text{uniformly} \Longrightarrow f(x) \ \text{is continuous}$. 
Hence this is impossible.
