12 For any $x>0$, there is always $f(x)=a-x-|\lg x| \leqslant 0$, then the range of values for $a$ is $\qquad$.
(12) (1) When $x \geqslant 1$, $|\lg x|=\lg x$. From $x+\lg x \geqslant 1$, we know $a \leqslant 1$;
(2) When $0 < x < 1$, $|\lg x|=-\lg x$. Let $f(x) = x - \lg x$, then $f'(x) = 1 - \frac{1}{x \ln 10}$. When $0 < x < \lg \mathrm{e}$, $f'(x) < 0$; when $x > \lg \mathrm{e}$, $f'(x) > 0$. Therefore, $f(x)$ reaches its minimum at $x = \lg \mathrm{e}$. Since $\lg \mathrm{e} < 1$ and $\ln 10 > 1$, we have
$$
\lg \mathrm{e} - \lg \lg \mathrm{e} = \lg \frac{\mathrm{e}}{\lg \mathrm{e}} = \lg (\mathrm{e} \cdot \ln 10) < \lg 10 = 1.
$$

Thus, the range of $a$ is $(-\infty, \lg \mathrm{e} - \lg \lg \mathrm{e}]$.