Answer
Suppose that $a\lt b$
$$
\begin{aligned} & \int_{-\infty}^{a} f(x) d x+\int_{a}^{\infty} f(x) d x \\
& \quad =\lim _{t \rightarrow-\infty} \int_{t}^{a} f(x) d x+\lim _{u \rightarrow \infty} \int_{a}^{u} f(x) d x \\ & \quad=\lim _{t \rightarrow-\infty} \int_{t}^{a} f(x) d x+\lim _{u \rightarrow \infty}\left[\int_{a}^{b} f(x) d x+\int_{b}^{u} f(x) d x\right] \\ &\quad =\lim _{t \rightarrow-\infty} \int_{t}^{a} f(x) d x+\int_{a}^{b} f(x) d x+\lim _{u \rightarrow \infty} \int_{b}^{u} f(x) d x \\ &\quad =\lim _{t \rightarrow-\infty}\left[\int_{t}^{a} f(x) d x+\int_{a}^{b} f(x) d x\right]+\int_{b}^{\infty} f(x) d x \\ & \quad =\lim _{t \rightarrow-\infty} \int_{t}^{b} f(x) d x+\int_{b}^{\infty} f(x) d x \\
& \quad=\int_{-\infty}^{b} f(x) d x+\int_{b}^{\infty} f(x) d x \end{aligned}
$$
Therefore,
$$
\int_{-\infty}^{a} f(x) d x+\int_{a}^{\infty} f(x) d x =\int_{-\infty}^{b} f(x) d x+\int_{b}^{\infty} f(x) d x
$$
Work Step by Step
Suppose that $a\lt b$
$$
\begin{aligned} & \int_{-\infty}^{a} f(x) d x+\int_{a}^{\infty} f(x) d x \\
& \quad =\lim _{t \rightarrow-\infty} \int_{t}^{a} f(x) d x+\lim _{u \rightarrow \infty} \int_{a}^{u} f(x) d x \\ & \quad=\lim _{t \rightarrow-\infty} \int_{t}^{a} f(x) d x+\lim _{u \rightarrow \infty}\left[\int_{a}^{b} f(x) d x+\int_{b}^{u} f(x) d x\right] \\ &\quad =\lim _{t \rightarrow-\infty} \int_{t}^{a} f(x) d x+\int_{a}^{b} f(x) d x+\lim _{u \rightarrow \infty} \int_{b}^{u} f(x) d x \\ &\quad =\lim _{t \rightarrow-\infty}\left[\int_{t}^{a} f(x) d x+\int_{a}^{b} f(x) d x\right]+\int_{b}^{\infty} f(x) d x \\ & \quad =\lim _{t \rightarrow-\infty} \int_{t}^{b} f(x) d x+\int_{b}^{\infty} f(x) d x \\
& \quad=\int_{-\infty}^{b} f(x) d x+\int_{b}^{\infty} f(x) d x \end{aligned}
$$
Therefore,
$$
\int_{-\infty}^{a} f(x) d x+\int_{a}^{\infty} f(x) d x =\int_{-\infty}^{b} f(x) d x+\int_{b}^{\infty} f(x) d x
$$
