Show that $f\in C([-1,1],\mathbb{R})$ and that $\int_{-1}^{1}f(x)\,dx=0$. Consider $X=C([-1,1],\mathbb{R})=\{f:[-1,1]\rightarrow \mathbb{R}\,:\,f \text{ is continuous}\}$ with the supremum metric defined by $$d(f,g)=\text{sup}_{x\in[-1,1]}|f(x)-g(x)|$$ for all $f,g \in X$. 
Show that if $(f_n)_{n=1}^{\infty}$ is a sequence in $X$ such that $\int_{-1}^{1}f_{n}(x)\,dx=0$ for all $n\in\mathbb{N}$, and that $f_n \rightarrow f$ as $n\rightarrow\infty$ in $(X,d)$ for some function $f$, then $$f\in X\,\,\,\,\,\,\, \text{ and } \,\,\,\, \int_{-1}^{1}f(x)\,dx=0.$$

I'm firstly a little confused about how to to show $f \in X$. Do I look at the usual $\epsilon-\delta$ definition for continuity of real functions (since isn't $X$ just all real, continuous functions $f:[-1,1]\rightarrow \mathbb{R}$), or more probably whether I need to consider $d$? Either way I've got to somehow get $f_n$ involved in that, and I guess I'll need to use the obvious integral inequality somewhere ($\left\lvert\int_a^b f(x)\,dx\right\rvert\leq\int_a^b\lvert f(x)\rvert dx$), but I'm not sure how to get the ball rolling.
 A: If $(f_n)_{n \in \mathbb{N}}$ converges in $(X,d)$ to some function $f$ that means:
$$\sup_{x \in [-1,1]} |f_n(x)-f(x)| \stackrel{n \rightarrow +\infty}\rightarrow 0$$
which implies $|f_n(x)-f(x)| \rightarrow 0$ for any $x \in [-1,1]$. This is called uniform convergence (because you are not fixing $x$ in your domain, as you would do for pointwise convergence) and trivially implies pointwise convergence (check!).
For $f$ continuous you would like to check that for any $\varepsilon > 0$ then you can find $\delta > 0$ such that $|f(x)-f(x_0)| < \varepsilon$ when $x \in (x_0-\delta,x_0+\delta)$. Then you can observe that by triangle inequality
$$|f(x)-f(x_0)| \leq |f(x)-f_n(x)|+|f_n(x)-f_n(x_0)|+|f_n(x_0)-f(x_0)|.$$
I let you conclude this part.
About the integral: we had
$$\forall \varepsilon > 0 \mbox{ }\exists N \in \mathbb{N}: |f_n(x)-f(x)| < \varepsilon \mbox{ when } n > N, \forall x \in [-1,1].$$
Then
$$-\varepsilon +f_n(x) < f(x) < f_n(x)+\varepsilon$$
hence if you integrate, you get $\int_{-1}^1 f(x)dx = 0$ (check!).
A: If a sequence of functions $f_n$ converges uniformly to  $f$ (i.e with the supremum norm)then $$\int_{-1}^1f_n \to \int_{-1}^1f$$
Also these functions are continuous on a closed bounded interval thus they are Riemman integrable.
And again from uniform convergence we have that a uniform limit of a sequence of continuous functions is continuous.
Now from uniqueness of limit we have that $\int_{-1}^1f=0$ because $\int_{-1}^1f_n=0,\forall n \in \Bbb{N}$
