TITLE: Integration over a metric space
QUESTION [0 upvotes]: Let $X$ be a metric space which is unbounded in the sense that $\text{diam}(X)<\infty$ with metric $d$. Let $u:X\to(0,\infty)$ be a positive measurable function such that for every ball $$B_R(0):=\{x\in X:d(x,0)<R\}.$$
Suppose
$$
\int_{B_R(0)}u\,dx\leq\frac{1}{R},
$$
for every $R>0$.
Then does it imply that 
$$
\int_{X}u\,dx=0?
$$
If $X=\mathbb{R}$, by letting $R\to\infty$ we get $\int_{\mathbb{R}}u\,dx=0$.
Can somebody please help me. Thanks.

REPLY [0 votes]: Assuming that $0$ stands for some element of the metric space and $dx$ is some Borel measure , the answer is YES. By Monotone Convergence Theorem $\int_X u dx =\lim \int_{B(0,R)} u dx \to 0$.
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