Integration over a metric space

Question

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.

Answer 1

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 $\int_X u dx =\lim \int_{B(0,R)} u dx \to 0$.