= Solution
For an <Lebesgue integrable function>[integrable] function $f:\mathbb R^n\to\mathbb C$, the <Lebesgue differentiation theorem> states that
$$
\lim_{r\downarrow0}\frac1{\lambda(B(x,r))}
\int_{B(x,r)}|f(y)-f(x)|\,d\lambda(y)=0
$$
for <almost everywhere> $x\in\mathbb R^n$. In particular,
$$
\lim_{r\downarrow0}\frac1{\lambda(B(x,r))}
\int_{B(x,r)}f(y)\,d\lambda(y)=f(x)
$$
at every <Lebesgue point> of $f$.
Almost every $x\in\mathbb R$ is a Lebesgue point of $g$. At such an $x$, for $t>0$,
$$
\left|\frac{G(x+t)-G(x)}t-g(x)\right|
\leq\frac1t\int_x^{x+t}|g(y)-g(x)|\,d\lambda(y)\longrightarrow0.
$$
For $t<0$, the same estimate over $[x+t,x]$ gives the identical <limit>. Therefore
$$
\boxed{G'(x)=g(x)}
$$
at every Lebesgue point of $g$, so $G$ is <differentiable function>[differentiable] $\lambda$-almost everywhere. This is the <differentiation of an indefinite Lebesgue integral>.
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