whether the set is dense? $X=C[0,1]$ with sup norm,
$S =\lbrace f\in X: \int _0^1f(t)dt \neq 0  \rbrace $ then how to decide whether S dense, nowhere dense ? 
The function which are outside S have integral $0$ so they are zero almost everywhere on $[0,1]$. Again all the positive$(f(x)>0$ $\forall x) $ and negative functions will be there in S. intuitively it is clear to me that this set is dense.
I have tried considering a function $f\to \int _0^1f$ so under this map $S$ is inverse image of $(-\infty,0) \cup (0,\infty)$. somehow i am not able to connect all this.. can anyone give some hint?
thank you.  
 A: First of:$\int_0 ^1 f = 0 $ does not implify $f \overset{\text{a.e.}}{=} 0$.
My attempt: Let $f\in C_\infty [0,1]$ if $\int_0 ^1 f \ne 0 $ then take the sequence $S \ni f_n = f$ obviously $f_n \rightarrow f$.
Otherwise, consider the sequence $f_n = f+ \frac{1}{n} $
Indeed $f\in C_\infty$ (as sum of such) and since $f \notin S$ we have: 
$$\int _0 ^1 f_n = \int _0 ^{1} f +\int _0 ^1 \frac{1}{n}=0+1/n \ne 0$$
That implies $f_n \in f$. We also have $\|f_n - f\|_\infty =\|\frac{1}{n}\|_\infty \rightarrow 0$.
Followup question: I'm interested to see if someone can approach this via functional analysis. considering the functional $T \in C^\text{#}_\infty$:
$Tf = \int_0^1 f$. We can see $T$ is bounded since $|Tf| \leq |\int_0^1 f | \leq \|f\|_\infty$ which implies the kernel is closed. We also know that $\mathrm{Codim} \ker T =1$, but i can't get further with this.
A: A functional analysis aproach:
Lema: If $V$ is a vector subspace of a normed space $X$, then $V^\circ=\emptyset$ or $V=X$ (here, $V^\circ$ is interior of $V$).
Proof: Suppose $V\neq X$ and $V^\circ\neq\emptyset$. Set $x_0\in V^\circ$. How $T:X\to X$ given by $Tx=x-x_0$ is continuos, we can suppose $x_0=0$. Now, there is an $\epsilon>0$ such that $B(0,\epsilon)\subseteq V$. Take $v\in X\backslash V$. Then, $v\neq 0$ and there is a non zero multiple of $v$ that is in $B(0,\epsilon)$, say $tv$. Then, $tv\in V$. Thus, every multiple of $tv$ belongs to $V$. In particular $v$. This contradiction shows that $V=X$ or $V^\circ=\emptyset$.
Now, let $T:C[0,1]\to C[0,1]$ given by $Tf=\int_0^1f(x)dx$. Note that $S=X\backslash\mathrm{Ker}(T)$. Then $\overline{S}=X\backslash\mathrm{Ker}(T)^\circ=X$ cause $\mathrm{Ker}(T)$ is a subspace of $X$ and it isn't $X$.
From that, $S$ is dense in $X$. Also, $S$ is an open set. Furthermore, note that we don't use that $T$ is continous. Then, if $T\neq 0$ is any linear operator, $\overline{X\backslash \mathrm{Ker}(T)}=X\backslash \mathrm{Ker}(T)^\circ=X$. Thus, $X\backslash\mathrm{Ker}(T)$ always is dense if $T$ isn't the null operator.
A: Hint :
$\displaystyle S=\left\{f\in X:\int_0^1f(t)\,dt \not=0\right\}=\left\{f\in X:\int_0^1f(t)\,dt<0\right\}\cup \left\{f\in X:\int_0^1f(t)\,dt>0\right\}$. 
What is $\bar{S}$ ?

 $\bar{S}=C[0,1]=X.$ and so ........

