9. Given the sequence $\left\{a_{n}\right\}$ with the sum of the first $n$ terms being $S_{n}, a_{1}=\frac{1}{5}$, and for any positive integers $m, n$, it holds that $a_{m+n}=a_{m} a_{n}$. If $S_{n}<a$ for any $n \in \mathbf{Z}_{+}$, then the minimum value of the real number $a$ is
9. $\frac{1}{4}$.

According to the problem, taking $m=1$ we get
$$
a_{n+1}=a_{1} a_{n}=\frac{1}{5} a_{n} \text {. }
$$

Also, $a_{1}=\frac{1}{5}$, so $\left\{a_{n}\right\}$ is a geometric sequence with the first term $\frac{1}{5}$ and common ratio $\frac{1}{5}$, i.e.,
$$
a_{n}=\frac{1}{5^{n}}\left(n \in \mathbf{Z}_{+}\right) .
$$

Therefore, $S_{n}=a_{1}+a_{2}+\cdots+a_{n}=\frac{1}{5}+\frac{1}{5^{2}}+\cdots+\frac{1}{5^{n}}$
$$
=\frac{1}{5} \times \frac{1-\frac{1}{5^{n}}}{1-\frac{1}{5}}=\frac{1}{4}\left(1-\frac{1}{5^{n}}\right) \text {. }
$$

Since for any $n \in \mathbf{Z}_{+}$, $S_{n}<a_{1}$, we have $a=\frac{1}{4}$.