Is there an analogue of the fundamental theorem of calculus for probability distributions without a density? Assume that $\mu$ is a probability measure on $\mathbb{R}$, and that we have the values of the integral $$A(s) = \int_0^s x \,d\mu(x) + \int_s^\infty d\mu(x).$$ If $\mu(x) = f(x) \, dx$, we can retrieve $f$, and hence $\mu$ by taking the derivative

$$A'(x) = xf(x) - f(x).$$

If $f$ is differentiable, the second derivative gives $A''(x) = f(x)$ and we now have $\mu$.

What if we don't know that it has a density function, is their any means of effectively retrieving the probability measure?
(Assuming the measure is non-atomic) for $h &gt; 0$ we have
$$ (s-1) \mu([s,s+h)) \le A(s+h) - A(s) = \int_s^{s+h} (x-1)\; d\mu(x) \le (s+h-1) \mu([s,s+h))$$

and thus if $s &gt; 1$,
$$  \frac{A(s+h)-A(s)}{s+h-1} \le  \mu([s,s+h)) \le \frac{A(s+h)-A(s)}{s-1}$$
If $s+h &lt; 1$, the inequalities are reversed.
By taking sufficiently fine partitions, we can obtain arbitrarily precise bounds on the measure of an interval.
(Assuming the measure is non-atomic) for $h &gt; 0$ we have
$$ (s-1) \mu([s,s+h)) \le A(s+h) - A(s) = \int_s^{s+h} (x-1)\; d\mu(x) \le (s+h-1) \mu([s,s+h))$$

and thus if $s &gt; 1$,
$$  \frac{A(s+h)-A(s)}{s+h-1} \le  \mu([s,s+h)) \le \frac{A(s+h)-A(s)}{s-1}$$
If $s+h &lt; 1$, the inequalities are reversed.
By taking sufficiently fine partitions, we can obtain arbitrarily precise bounds on the measure of an interval.