# Extended Riemann integrability of a non-negative function implies Lebesgue integrability?

Let $f$ be a bounded function on a finite interval $[a, b]$ of the real line. If $f$ is Riemann integrable, we denote its Riemann integral by $\mathcal{R}(f , [a, b])$. It is well known that $f$ is Lebesgue integrable and $\mathcal{R}(f , [a, b]) = \int_{a}^{b} f dt$.

I would like to generalize this result in the following way.

Definition 1 Let $[a, b]$ be a finite interval of the real line. Let $f\colon [a, b] \rightarrow [0, \infty]$ be an extended real valued function. Suppose $f$ is Riemann integrable on every finite interval $[x, y]$, where $a< x < y < b$. We denote $\lim_{x\rightarrow a, y\rightarrow b, a < x < y < b} \mathcal{R}(f , [x, y])$ by $\mathcal{R}(f , [a, b])$ and we say $\mathcal{R}(f , [a, b])$ is definable. Note that $\mathcal{R}(f , [a, b])$ can be $\infty$. Note also that this definition coincides with the usual Riemann integral when $f$ is Riemann integrable on $[a, b]$.

Definition 2 Let $[a, b]$ be a finite interval of the real line. Let $f\colon [a, b] \rightarrow [0, \infty]$ be an extended real valued function. Let $a = t_0 < t_1 <\cdots < t_{k-1} < t_k = b$ be a partition of $[a, b]$. Suppose $\mathcal{R}(f , [t_{i-1}, t_i])$ is definable for $i = 1,2,\dots,k$. Then we denote $\sum_{i=1}^{k} \mathcal{R}(f , [t_{i-1}, t_i])$ by $\mathcal{R}(f , [a, b])$ and we say $\mathcal{R}(f , [a, b])$ is definable. If $\mathcal{R}(f , [a, b])$ is definable and $\mathcal{R}(f , [a, b]) < \infty$, we say $f$ is extended Riemann integrable and we say $\mathcal{R}(f , [a, \infty))$ is the extended Riemann integral of $f$.

Definition 3 Let $[a, \infty)$ be an interval of the real line, where $-\infty < a < \infty$. Let $f\colon [a, \infty) \rightarrow [0, \infty]$ be an extended real valued function. Suppose $\mathcal{R}(f , [a, b])$ is definable for every $b$ such that $a < b < \infty$. We denote $\lim_{b\rightarrow \infty} \mathcal{R}(f , [a, b])$ by $\mathcal{R}(f , [a, \infty))$ and we say $\mathcal{R}(f , [a, \infty))$ is definable. If $\mathcal{R}(f , [a, \infty))$ is definable and $\mathcal{R}(f , [a, \infty)) < \infty$, we say $f$ is extended Riemann integrable and we say $\mathcal{R}(f , [a, \infty))$ is the extended Riemann integral of $f$.

My question Is the following proposition true? If yes, how would you prove this?

Proposition Let $[a, \infty)$ be an interval of the real line, where $-\infty < a < \infty$. Let $f\colon [a, \infty) \rightarrow [0, \infty]$ be an extended real valued function. Suppose $\mathcal{R}(f , [a, \infty))$ is definable. Then $f$ is Lebesgue measurable and $\mathcal{R}(f , [a, \infty)) = \int_{a}^{\infty} f dt$. In particular, if $f$ is extended Riemann integrable, then $f$ is Lebesgue integrable.

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