Finding the Cardinality of a set. Let $A$ be the set of all continuous functions $f :[0,1] → [0,∞)$ satisfying the following condition:
$\int_{0}^{x} f(t) dt \geq f(x) $ for all $ x \in [0,1]$. 
Find the cardinality of the set $A$ . 
I have no idea how to procede . Please help. 
 A: Hint: how many continuous functions $f:[0,\,1]\mapsto [0,\,\infty)$ exist, satisfying the last condition or otherwise? (You can get an upper bound from the fact $f$ is specified by its values on $\Bbb Q\cap [0,\,1]$, and this upper bound is also an obvious lower bound by constructing a specific family of solutions.) And can you show this cardinality also lower-bounds $|A|$?
A: Clearly the identically zero function ($f(x)=0$ for all $x$) belongs to $A$.
I will prove that $A$ has no other element.
Let $f \in A$. Then, for all $x \in [0,1]$
$$\int_0^x f(t) \ \mathrm d t \ge f(x)$$
integrating on $x$ you have
$$\int_0^1 \int_0^x f(t) \ \mathrm d t \ \mathrm d x \ge \int_0^1 f(x) \ \mathrm d x$$
The first integral can be manipulated into
$$\int_0^1 \int_0^x f(t) \ \mathrm d t \ \mathrm d x =
\int_0^1 \int_t^1 f(t) \ \mathrm d x \ \mathrm d t = \int_0^1 (1-t) f(t) \ \mathrm d t = \int_0^1 f(t) \ \mathrm d t - \int_0^1 tf(t) \ \mathrm d t$$
Thus you have
$$\int_0^1 f(t) \ \mathrm d t - \int_0^1 tf(t) \ \mathrm d t \ge \int_0^1 f(x) \ \mathrm d x$$
i.e.
$$\int_0^1 t f(t) \ \mathrm d t \le 0$$
Since $tf(t)$ is continuous and non-negative, its integral must be non-negative. Thus you have
$$0 \le \int_0^1 t f(t) \ \mathrm d t \le 0$$
which implies that $f(t)$ is identically zero.
