Krdytkyu

I want to solve the following:

Let $ f:(a,b)tomathbb{R}$ continous such that $f(x)ge 0 $ for all $xin(a,b)$. Show that $f$ is integrable iff $displaystyle lim_{varepsilonto0} int_{[a+varepsilon,b-varepsilon]}f$ exists.

My attempt:

$Leftarrow]$ I want invoke the following proposition:

If $A$ is open and bounded, and $f:Atomathbb{R}$ is bounded and its set of discontinuities is measure zero, then $f$ is integrable.

And we have that $(a,b)$ is bounded by the one dimensional rectangle $[a,b]$ and since $f$ is continous we have that it is bounded in $(a,b)$, but the thing is that this argument does not need the limit. Can you help me fix this please?

If $f$ is integrable then we have that:

$$sum_{phi in F} phi f to int_{(a,b)} f = displaystyle lim_{varepsilonto0} int_{[a+varepsilon,b-varepsilon]}f$$

But I think this is a little bit trivial and I think I am wrong. Can you help me verify this, and if it is wrong, can you help me fix the mistakes please?

Thanks a lot in advance :)

The key is that $f(x) geq 0$. If you don't have that, it isn't true:

$int_{-1}^1 frac{x}{1-x^2}, dx$ does not exist (each asymptote goes like $1/x$ near $0$), but $int_{-1+epsilon}^{1-epsilon} frac{x}{1-x^2}, dx = 0$ for all $0 < epsilon < 1$ because the function is odd.

Note that you can't assume $f$ is bounded, or that it extends continuously to $[a,b]$. The usual definition of the Riemann integral only applies to bounded functions on closed intervals, so in this case we're looking at an integral which is potentially improper at $a$ and at $b$. We do have that $f$ is continuous on $(a,b)$, so it is integrable on any proper closed subinterval of $(a,b)$.

One definition of $int_a^b f(x), dx$ for integrals which are improper at $a$ and at $b$ (only) would be, choosing some $c in (a,b)$:

$int_a^b f(x), dx = lim_{trightarrow a^+} int_t^c f(x), dx$ + $lim_{s rightarrow b^-} int_c^s f(x), dx$ $qquad(*)$

The claim is that each of those limits exists provided that $f(x) geq 0$ and $lim_{epsilonrightarrow 0^+} int_{a+epsilon}^{b-epsilon} f(x), dx$ exists. Call that integral $I(epsilon) = int_{a+epsilon}^{b-epsilon} f(x), dx$.

Because $f(x) geq 0$, we have that $I(epsilon)$ is increasing as $epsilon$ decreases. Therefore the limit $lim_{epsilonrightarrow 0^+} I(epsilon)$ exists if and only if the set of values ${I(epsilon): epsilon > 0}$ is bounded, in which case the limit is the supremum of these values.

It then quickly follows that if this is true, then each of the limits above in equation $(*)$ exist because they, too, are bounded and increase as $t$ decreases and as $s$ increases.

I don't have Spivak on hand, and am not certain what definitions he (and hence you) are using, but you should be able to give a proof roughly equivalent to the above using whatever definitions you have.