If $f:[0,1]\to[0,\infty)$ is Riemann-Integrable on every closed subinterval of $(0,1]$, Is it possible that $f$ is not Lebesgue-Integrable?

Question

If $f\colon[0,1]\to[0,\infty)$ is Riemann-Integrable on every closed subinterval of $(0,1]$, Is it possible that $f$ is not Lebesgue-Integrable ?

According to a lemma(which has to be proven):

$f$ is Lebesgue-Integrable $\Leftrightarrow \lim_{\epsilon\downarrow0}\int_\epsilon^1f(x)\, dx$ exists in $\mathbb R$

then we have to assume that the limit doesn't exist, but is this not a contradiciton to the Riemann-Integrability ?

Answer 1

No, its not a contradiction. That limit existing means that the improper Riemann integral exists. Take for example $f(x)=1/\sqrt{x}$.

$f$ is not Riemann integrable on $[0,1]$ (it is not bounded), but it is improperly Riemann integrable in the sense that $\lim_{\epsilon\rightarrow 0} \int_\epsilon^1 f(x) dx$ exists (and therefore by your Lemma also Lebesgue integrable).

Regarding your initial statement: Yes it is possible, just give $f$ a non-integrable singularity at $0$, for example, $f(x)=1/x^2$ works. It is Riemann integrable on all finite closed subintervals of $(\epsilon, \infty)$ but not Lebesgue integrable at $0$.

Answer 2

Another example: $$ f(x) = \frac{1}{x^2}\sin\frac{1}{x},\qquad x \in (0,1] $$ Certainly $f$ is Riemenn integrable on $[a,1]$ for any $a>0$. But $$ \int_0^1 |f(x)|\,dx = +\infty $$ so $f$ is not Lebesgue integrable on $[0,1]$.

added

Even better $$ f(x) = \frac{1}{x}\sin\frac{1}{x},\qquad x \in (0,1] $$ Now the improper Riemann integral converges, but the Lebesgue integral still does not exist. This is essentially the fact that $\sum_{n=1}^\infty (-1)^n/n$ converges conditionally but not absolutely.