Inequality and integration Is the following true?
If $$\int_{0}^{x}f(t)\,dt \leq \int_{0}^{x} c \,dt =cx $$
for all $x>0$, $x$ is real number, and $c$ is some fixed constant,
then
$$f(t) \leq c$$ for all $t>0$?
EDIT: I should said that $f(t)$ is positive function on $t>0$, and $f(t_{1}+t_{2})\geq f(t_{1})+f(t_{2})$, for all $t_{1},t_{2}>0$ if this helps!
 A: With the condition provided in the "EDIT" section, the answer is: yes, it is true. Below is the proof.
Since $\int_0^x f(t)dt\leq \int_0^x cdt$, for all $x > 0$, we have $\int_0^x(c - f(t))dt\geq 0$ for all $x > 0$. It will suffice, of course, to prove from the last inequality (and the condition that for all $t_1, t_2 > 0$, $f(t_1 + t_2)\geq f(t_1) + f(t_2)$) that $c - f(x)\geq 0$ for all $x > 0$.
Due to lack of further information, I am assuming that we're dealing with Riemann integrals, and $f$ is Riemann integrable. 
Since for all $x, \epsilon > 0$, $f(x + \epsilon) \geq f(x) + f(\epsilon)$, and $f(\epsilon) > 0$ (since $f$ is positive on the positive Real half-line), we know that $f$ is nondecreasing. Now let us assume that for some $x_0 > 0$, $f(x_0) > c$. Say $f(x_0) - c = \delta > 0$. Then for all $x > x_0$, we must have $f(x) - c \geq \delta\implies f(x)\geq c + \delta$. Therefore we obtain, for any $x > x_0$,
$$
\int_{x_0}^x f(t)dt \geq \int_{x_0}^x c + \delta dt = (c + \delta)(x - x_0).
$$
On the other hand, say
$$
cx_0 - \int_0^{x_0}f(t)dt = \eta.
$$
Then if $x$ is sufficiently large, we have, from above, $(c + \delta)(x - x_0) > \eta$. In other words, for $x$ sufficiently large, we get
$$
\int_0^x f(t)dt > cx,
$$
contradicting the hypothesis.
Note that no extra assumption of continuity of $f$ is needed. Although if we assume that $f$ is Riemann integrable, then $f$ has only countably many discontinuities. 
The same argument works to establish analogous result in the case of Lebesgue integrability.
