# Existence of Locally (Lebesgue-)Integrable Function

Given a locally integrable function $$f: \mathbb R_{\geq0} \rightarrow \mathbb R_{\geq0}$$, I wonder whether there exists an equivalent function that operates at a certain capacity $$\nu\in\mathbb R_{>0}$$.

More specifically, I would like to show the existence of a (measurable) function $$g: \mathbb R_{\geq 0} \rightarrow \mathbb R_{\geq0}$$ that fulfills the following two conditions for almost all $$t\in\mathbb R_{\geq0}$$:

(1) $$g(t)\leq \nu$$

(2) $$g(t) = \begin{cases} \nu, &\text{if } q(t) := \int_{0}^t f - g \, \mathrm{d}\lambda > 0,\\ f(t), &\text{otherwise.} \end{cases}$$

It seems quite obvious, that such a function should exist, as the condition "$$\int_{0}^t f - g \, \mathrm{d}\lambda > 0$$" to determine $$g(t)$$ only depends on "the past" of $$g$$, i.e. only on $$g\vert_{(0,t)}$$.

Nevertheless, I lack a formal proof of the existence of $$g$$.

One idea I had was to first show the statement for simple functions, i.e. $$f = \sum_{i=1}^k a_k \mathbf{1}_{A_k}$$ for measurable sets $$A_k$$ and $$a_k > 0$$ and then use an approximation using only simple functions. Nevertheless, I got stuck here. Maybe someone else has an idea or a proof sketch? Is it even impossible to do for some integrable functions $$f$$?

It is worth mentioning that the origin of the question lies in queue dynamics of dynamic flows (also called flows over time) with the so-called Vickrey's deterministic fluid queuing. More specifically, I found an answer in the paper [1, Section 2.2] of Cominetti, Correa und Larré. I worked out the relevant proofs in the following.

It has proven beneficial to first analyze the uniqueness of the queue function $$q(t):= \int_0^t f - g \,\mathrm{d}\lambda$$.

Claim 1. Given a function $$g$$ fulfilling (1) and (2) a.e., the queue function is of the form $$q(t) = \max_{u\in[0,t]} \int_u^t f - \nu \, \mathrm d\lambda.$$ Proof. We first show, that $$q$$ is never negative. Assume the contrary and let $$t>0$$ fulfill $$q(t) < 0$$. Choose $$u^* := \max\{u\leq t \,\vert\, q(u) = 0 \}$$ to be the latest time before $$t$$ at which the queue was zero. By the continuity of $$q$$ we know that $$q$$ is strictly negative on $$(u^*, t]$$. Thus using property (2) we get $$q(t) = \int_0^t f - g \,\mathrm d\lambda = \int_0^{u^*} f - g \,\mathrm d\lambda + \int_{u^*}^t f - f \,\mathrm d\lambda = q(u^{*})=0,$$ a contradiction.

Now, we focus on the statement of the claim. Let $$t\in \mathbb R$$ and let us define $$u^* := \max\{u\leq t \,\vert\, q(u) = 0 \}$$ again to be the latest time before $$t$$ at which the queue is empty. Then by definition we have $$q(u) > 0$$ for all $$u\in(u^*, t]$$ and hence (2) implies $$g(u) = \nu$$ for almost all $$u\in(u^*, t]$$. This shows $$q(t) = \int_0^{u^*} f - g \,\mathrm d\lambda + \int_{u^*}^t f - g \,\mathrm{d}\lambda = q(u^*) - \int_{u^*}^t f - \nu \,\mathrm{d}\lambda = \int_{u^*}^t f - \nu \,\mathrm{d}\lambda.$$

It remains to show, that $$\int_u^t f - \nu \,\mathrm d\lambda \leq q(t)$$ holds for any other $$u\in[0,t]$$. If $$u\leq u^*$$, then (1) implies $$\int_u^t f - \nu \,\mathrm d\lambda\leq \int_u^t f - g\,\mathrm d\lambda = q(t) - q(u) \leq q(t).$$ If $$u > u^*$$, we use (2) and get $$\int_u^t f - \nu \,\mathrm d\lambda = q(t) - \int_{u^*}^{u}f - \nu \,\mathrm d\lambda - q(u^*) = q(t) - q(u) \leq q(t).$$ $$\square$$

Note: We haven't used that $$\nu$$ is constant. Instead it could be any locally integrable function.

Claim 2. The function $$q(t) := \max_{u\in[0,t]} \int_u^t f - \nu \, \mathrm d\lambda$$ is almost everywhere differentiable with $$q'(t) = \begin{cases} f(t) - \nu, &\text{if } q(t) > 0,\\ 0, &\text{otherwise}. \end{cases}$$ Moreover, we have $$f(t) \leq \nu(t)$$ for almost all $$t$$ with $$q(t)=0$$.

I refer to this answer for a worked out proof. The "Moreover"-part is proven here.

Note: We could replace $$\nu$$ by any non-negative locally integrable function.

Final Claim. Given a locally integrable function $$f:\mathbb R_{\geq 0}\rightarrow \mathbb R_{\geq0}$$, there exists a locally integrable $$g: \mathbb R_{\geq 0}\rightarrow \mathbb R_{\geq0}$$ fulfilling (1) and (2) a.e.. Moreover, $$g$$ is unique up to a null set.

Proof. We first define $$q(t) := \max_{u\in[0,t]} \int_u^t f - \nu \,\mathrm d \lambda$$. We define $$g(t) := f(t) - q'(t)$$ wherever $$q'(t)$$ is defined, and $$g(t):=0$$ elsewhere. Then, we have $$q(t) = \int_0^t f - g \,\mathrm d\lambda$$ and $$g$$ automatically fulfills (2) a.e. by Claim 2.

It remains to show, that $$g(t)\leq \nu$$ almost everywhere. For $$q(t) > 0$$, this is given by (2) already. For $$q(t)=0$$ this follows from Claim 2.

Now assume there is another function $$h$$ fulfilling (1) and (2) almost everywhere. By Claim 1 we have $$\int_0^t f - g \,\mathrm d\lambda = \max_{u\in[0,t]} \int_u^t f -\nu \,\mathrm d\lambda = \int_0^t f -h \,\mathrm d\lambda.$$ for any $$t\geq 0$$. Hence $$g$$ and $$h$$ coincide almost everywhere.

$$\square$$

Note: We haven't used that $$\nu$$ is constant. Instead it could be any locally integrable function.

References

 "Dynamic Equilibria in Fluid Queuing Networks" by Roberto Cominetti, José R. Correa, Omar Larré. Available at arXiv:1401.6914 [math.OC].