Is there a lower bound for $\int_{B_{r}}f$ when $f$ is a positive function? Let $f$ be a positive Lebesgue integrable function on a ball $B_{r}\in\mathbb{R}^n$. I'm looking for a positive constant $c$ depending only on $f$ that satisfies
$$c\omega_nr^n\leq\int_{B_r}f$$
where $\omega_n$ is the volume of the $n$-dimensional unit ball. 
I'm also wondering if the similar result holds for a nonnegative integrable function $f$ and a nonnegative constant $c=c(f)$. Of course, $c=0$ satisfies the inequality in the this case but I need a positive constant $c$ unless $f\equiv 0$ a.e. on $B_r$. 
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I have asked these questions as I thought that the existence of such a constant will help to fill out the gap of the proof of Proposition 2.27(Alexandroff's Maximum Principle of Quasilinear Equations) in "Elliptic Partial Differential Equations" by Q. Han ad F. Lin. 
In the proof it derives an inequality
$$\int_{B_M}g^n\leq\int_{\Omega}h^n<\left(\int_{\mathbb{R}^n}g^n\right)$$
where $g\in L^n_{loc}(\mathbb{R}^n)$ and $h\in L^n(\Omega)$ are nonnegative functions and $M$ a positive constant. (The second inequality is just an assumption of the proposition.) Then the authors conclude with no further explanation that there exists a positive constant $C$ which depends only on $g$ and $h$ such that 
$$M\leq C.$$
I think that in order to derive the second inequality from the first one we need a bound which I have suggested. Thanks for your help.
 A: Both the essential infimum ${\rm essinf}_{B_r}(f)$ and the mean value $\frac{1}{\omega_n r^n}\int_{B_r} f$ will work, as pointed out in the answer above.
Note, also, that both of these depend on both $f$ and the domain (and hence on $r$), but they are sharp bounds as long as there's no further information on $f$. They may be somewhat unsatisfactory - especially the mean value which leads into a tautology. In that case, you should ask the question in a more stringent manner.
Can a bound $c$ only depending on $f$ (not on $r$) be found? Well, ${\rm essinf}_{\Omega}(f)$ is the obvious choice, with $\Omega \supset B_r$ the domain of definition of $f$. This, though, may not be positive even if ${\rm essinf}_{B_r}(f)$ is: $\Omega \supset B_r \Rightarrow {\rm essinf}_{\Omega}(f) \le {\rm essinf}_{B_r}(f)$. It would help knowing that there exists a ball $B_\rho \subset B_r$ such that ${\rm essinf}_{B_\rho}(f) > 0$, especially if your domain is not infinite. Do you have any such information? What is the context for this problem anyway?
A: Regarding the original problem you edited in: just take $C$ to be the supremum of $M$ such that the first inequality is true. Since $\int_{B_M} g^n \to \int_{\mathbb{R}^3} g^n$ as $M \to \infty$ and $\int_{\Omega} h^n < \int_{\mathbb{R}^3} g^n$, we know $\int_{B_M} g^n$ must eventually stay greater than $\int_{\Omega} h^n$, and thus the supremum must be finite.
A: If $f>0$ on $B_r$, and $c \le f(x)$ for all $x\in B_r$, we know that the volume of $B_r$ is $\omega_n r^n$, so we get
$$
c\,\omega_n r^n = c\,\operatorname{Vol}(B_r) = c\int_{B_r} 1\,\mathrm dx = \int_{B_r} c\,\mathrm dx \le \int_{B_r} f(x)\,\mathrm dx.
$$
Thus, $c(r,f)=\inf_{x\in B_r} f(x)$ satisfies your inequality.
However, as @martini pointed out, this infimum might be $0$.
Instead, consider the mean value of $f$ on $B_r$, i.e. $\overline f = \frac{\int_{B_r} f}{\operatorname{Vol}(B_n)}$. We easily see that
$$\int_{B_r} \overline f = \overline f\,\omega_n r^n = \int_{B_r} f.$$
Thus, $c=\overline f$ also satisfies your inequality, where $\overline f = 0$ if and only if $f\equiv 0$ almost everywhere on $B_r$, so $f>0$ is certainly sufficient for $\overline f>0$.
Of course this is an obvious bound you immediatly get from dividing the inequality by $\omega_n r^n$ in the first place.
For non-negative $f$ you will not get a bound, since $\int_{B_r} f$ might be $0$ in this case. Consider for example
$$
f(x) = 
\begin{cases}
0,& \text{if $x\neq 0$,}\\
1,& \text{if $x=0$.}
\end{cases}
$$
