4. Given $a+\lg a=10, b+10^{b}=10$. Then $a+b$
4. 10.

Thought 1: From the given information,
$$
\begin{array}{l}
a=10^{10-a}, \\
10-b=10^{b} .
\end{array}
$$

Subtracting, we get $10-a-b=10^{b}-10^{10-a}$.
If $10-a-b>0$, then $10-a>b$. By the monotonicity of the exponential function, we get $10^{10-a}>10^{b}$. Substituting into (1), we get
$$
010-a-b=10^{b}-10^{10-a}>0 \text{. }
$$

Contradiction.
Therefore, $10-a-b=0$, which means $a+b=10$.
Thought 2: The geometric meaning of this problem is: $a$ is the x-coordinate of the intersection point $A$ of $y=\lg x$ and $y=10-x$, and $b$ is the x-coordinate of the intersection point $B$ of $y=10^{x}$ and $y=10-x$. By the symmetry of the graphs of two inverse functions, $A$ and $B$ are symmetric about the line $y=x$. Solving the system of equations
$$
\left\{\begin{array}{l}
y=x \\
y=10-x
\end{array}\right.
$$

we get $C(5,5)$. Since $C$ is the midpoint of $A$ and $B$, we have $\frac{a+b}{2}=5$, which means $a+b=10$.