A function of two variables and an integral

Question

Find nontrivial (not constant zero) real functions $P$ of two arguments (or prove that there is no such function), such that

$$ P (T, s) = \int_T^{T + s} P (x, \mathrm{d} x) $$ where by definition $$ \int_T^{T + s} P (x, \mathrm{d} x) = \lim_{N \rightarrow + \infty} \sum_{i = 0}^{N - 1} P \left( T + \frac{i s}{N}, \frac{s}{N} \right). $$

You can assume that $P$ is continuous and otherwise well behaves.

$T$ is always nonnegative and $s$ is always positive. $P$ is nonnegative.

(I prefer not to disclose the details of where I get this problem from. It is from economics. But I can say that you'll help to the world if you find this function.)

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Another (possibly simpler) variant of this problem: Find $P$ such that $$P(T,a+b) = P(T,a) + P(T+a,b)$$ for nonnegative $T$ and positive $a$, $b$.

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One of the solutions is $P(T,s) = k s$ (where $k=\mathrm{const}$). This solution does not fit into the actual problem domain. Are there other solutions?

Answer 1

$P(T,s)$ must be linear for the second argument, otherwise the limit is not well defined.

So we can assume $P(T,s)=f(T) k s$ for a fixed $k$. Then

$f (T) k (a + b) = f (T) k a + f (T + a) k b$;

$f (T) (a + b) = f (T) a + f (T + a) b$;

$f (T) b = f (T + a) b$;

$f (T) = f (T + a)$ that is $f(T)=k_2$ is a constant function. Thus $P(T,s)=k_2 k s$ that is $P$ is just a linear function of $s$. There are no other solutions.