# A basic question about integration

I have read that if $f$ is integrable, it is bounded. But consider $f=x^{-\frac 12}$, it is integrable in $[0,1]$, but it is not bounded. Can you explain？

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$x^{-\frac1{2}}$ is not integrable in $[0, 1]$, because it's not defined at $x = 0$. – Javier Feb 19 '12 at 14:13
@JavierBadia if I define f=0 when x=0,it doesn't matter – 89085731 Feb 19 '12 at 14:15
We have a positive function, and we look at "the area under the graph of $f$", the fact that $f$ is bounded or not doesn't matter if the area on $[a,1]$ doesn't "grow to fast" if $a$ goes to $0$. – Davide Giraudo Feb 19 '12 at 14:20
@Gingerjin: the limit when $x$ goes to $0$ is $\infty$, not $0$, so it doesn't make sense to define it as such. It would not be continuous. – Javier Feb 19 '12 at 14:25
@JavierBadia it doesn't matter whether it is continuous or not. – 89085731 Feb 19 '12 at 14:27

$\int_0^1 x^{-1/2}\,dx$ is called an improper integral.

$f(x)=x^{-1/2}$ is not (Riemann) integrable over $[0,1]$, as you said since this function is not bounded over $[0,1]$ (and not defined at $0$; but let's say $f(0)=0$ in the following). This follows from the definition of the Riemann integral in terms of Riemann sums:

$f$ is Riemann integrable (henceforth just "integrable") over $[a,b]$ and $\int_a^b f(x)\,dx=L$ if for any $\epsilon>0$ there is a $\delta>0$ so that for any Riemann sum $R_f$ over $[a,b]$ corresponding to a partition with mesh size at most $\delta$, $$|L - R_f|<\epsilon$$

For a non-bounded function over $[a,b]$, one can find a partition of $[a,b]$ of arbitrarily small mesh size and a corresponding Riemann sum which is as large as desired in absolute value (for $f(x)=x^{-1/2}$, take the first subinterval of the partition to be $[0,\delta]$ for $\delta$ small, and choose a test point $x_1$ in this interval so that the height of the rectangle corresponding to this subinterval is large). This implies that the function is not integrable over $[a,b]$.

So unbounded functions are not integrable, a fact that crops out from the very definition of integrability.

But, sometimes, an unbounded function can be thought of as being "integrable". For example, the area under the graph of $f(x)=x^{-1/2}$ over $[0,1]$ can be seen to be finite:

The function $f(x)=x^{-1/2}$ is integrable over $[\delta,1]$ for any $1\ge \delta>0$. Moreover, $\lim_{\delta\rightarrow 0^+}\int_\delta^1 x^{-1/2}\, dx$ exists. It seems plausible to define the area under the graph of $f(x)=x^{-1/2}$ over $[0,1]$ as the value of this limit.

For this reason, we say (or should say) $f(x)=x^{-1/2}$ is improperly integrable over $[0,1]$.

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For a function $f$ to be Riemann integrable on a given interval, indeed $f$ must be bounded. Otherwise, no upper Darboux sum is finite.

Lebesgue integrability is different and does not require the function to be bounded, for example $f:x\mapsto x^{-1/4}$ is Lebesgue integrable on $(0,1)$ (and its integral is $4/3$). A simple way to see that this function is Lebesgue integrable is to consider $g:x\mapsto 1+\sum\limits_{n\geqslant1}[0\lt x\leqslant 1/n^4]$. Then, the function $f$ is continuous, $0\leqslant f\leqslant g$ on $(0,1)$, and the Lebesgue integral of $g$ is $1+\sum\limits_{n\geqslant1}1/n^4$, which is finite, hence $f$ is Lebesgue integrable.

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Absolutely: the function $x\mapsto x^{-1/2}$ is not Riemann integrable on $[0,1]$ because it is not bounded on this interval. – Did Feb 19 '12 at 14:42
Are you sure $g$ is a simple function? It doesn't look like a finite sum of characteristic functions to me. – Rob Arthan Feb 20 '12 at 23:30
@Rob: You are right, I was too hasty. Thanks. – Did Feb 21 '12 at 7:02

Everything depends on what integral we are talking about. Is it Riemann integral, Lebesgue integral? For Riemann integral we have a definition with Riemann sums. If the function isn't bounded, then the value of any Riemann sum can be arbitrarily big, so they don't converge to a finite value and the function isn't integrable. However people noticed the problem and if there is a singularity at a certain point (like $0$ here) we can compute the limit of integrals $\int_\varepsilon^1 f$ and call it the integral over $[0,1]$. It's called a "principal value".

For Lebesgue integral there are no such limitations (but it doesn't mean that it is always better than Riemann integral).

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As has already been said the property that only bounded functions are integrable depends on which theory of integration you are using. In the Riemann theory, unbounded functions have to be treated specially as "improper integrals", i.e., limits of integrals of bounded functions.

The gauge integral of Henstock and Kurzweil is a generalisation of the Riemann integral that avoids the need for the notion of improper integral. The gauge integral is to the Riemann integral as uniform continuity is to continuity: in the Riemann integral the strips in a partition all have the same width, while in the gauge integral the widths of the strips can vary, so that you can control the impact of local bad behaviour. This means that integrals like $\int_0^1 x^{-\frac{1}{2}}dx$ don't need any special treatment - the additional limiting process needed to give the improper integral in the Riemann theory is superfluous.

The gauge integral has other advantages too. E.g., unlike both the Riemann and the Lebesgue theory, the fundamental theorem of the calculus for the gauge integral does not need to assume that the integrand is integrable. It is a great shame that the gauge integral is not more widely taught.

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