8. Given real numbers $a, b$ satisfy
$$
a+\lg a=10, b+10^{b}=10 \text {. }
$$

Then $\lg (a+b)=$ $\qquad$
8. 1 .

Since $\lg a=10-a, 10^{b}=10-b$, therefore, $a$ is the x-coordinate of the intersection point of $y=\lg x$ and $y=10-x$, and $b$ is the y-coordinate of the intersection point of $y=10^{x}$ and $y=10-x$.

Also, $y=\lg x$ and $y=10^{x}$ are symmetric about the line $y=x$, and $y=10-x$ is symmetric about the line $y=x$, so the two intersection points are symmetric about the line $y=x$.
Thus $10-b=a \Rightarrow \lg (a+b)=1$.