One of the first surprises students encounter in measure theory is that unbounded functions can still be integrable.

At first this seems counterintuitive. If a function becomes arbitrarily large, shouldn’t its area also become arbitrarily large?

Surprisingly, the answer is no.

The key question is not whether a function is unbounded, but whether the total area accumulated under its graph is finite.

This distinction leads directly to one of the central ideas of Lebesgue integration.

The Lebesgue Integrability Criterion

The fundamental theorem governing integrability is the following.

Lebesgue Integrability Criterion

A measurable function (f) is Lebesgue integrable if and only if:

$$\int |f|,d\mu<\infty$$

In other words, a function is integrable precisely when the integral of its absolute value is finite.

Notice that boundedness does not appear anywhere in this criterion.

A function may become infinitely large and still be integrable.

An Unbounded Function That Is Integrable

Consider:

$$f(x)=\frac{1}{\sqrt{x}}$$

on the interval:

$$[0,1]$$

As $x\to0^+$,

$$\frac{1}{\sqrt{x}}\to\infty$$

so the function is clearly unbounded.

To determine whether it is integrable, we compute:

$$\int_0^1 \frac{1}{\sqrt{x}},dx$$

Using elementary calculus:

$$\int_0^1 \frac{1}{\sqrt{x}},dx=\lim_{\varepsilon\to0^+}\int_\varepsilon^1 x^{-1/2},dx$$

and

$$\int_\varepsilon^1 x^{-1/2},dx=\left[2\sqrt{x}\right]_\varepsilon^1=2-2\sqrt{\varepsilon}$$

Taking the limit:

$$\lim_{\varepsilon\to0^+}(2-2\sqrt{\varepsilon})=2$$

Therefore:

$$\int_0^1 \frac{1}{\sqrt{x}},dx=2$$

Since the integral is finite, the function is Lebesgue integrable.

An Unbounded Function That Is Not Integrable

Now consider:

$$f(x)=\frac{1}{x}$$

on ((0,1]).

Again:

$$\frac{1}{x}\to\infty$$

as $x\to0^+$.

To test integrability we compute:

$$\int_0^1 \frac{1}{x},dx$$

This equals:

$$\lim_{\varepsilon\to0^+}\int_\varepsilon^1 \frac{1}{x},dx$$

and

$$\int_\varepsilon^1 \frac{1}{x},dx=\left[\ln x\right]_\varepsilon^1=-\ln(\varepsilon)$$

As $\varepsilon\to0^+$,

$$-\ln(\varepsilon)\to\infty$$

Thus:

$$\int_0^1 \frac{1}{x},dx=\infty$$

The function is therefore not Lebesgue integrable.

The p-Integral Test

These examples are instances of a more general theorem.

p-Integral Test

The improper integral:

$$\int_0^1 \frac{1}{x^p},dx$$

converges if and only if:

$$p<1$$

and diverges if and only if:

$$p\ge1$$

This theorem immediately explains the previous examples.

For:

$$\frac{1}{\sqrt{x}}=\frac{1}{x^{1/2}}$$

we have:

$$p=\frac12<1$$

so the integral converges.

For:

$$\frac{1}{x}=\frac{1}{x^1}$$

we have:

$$p=1$$

so the integral diverges.

The function (1/x) lies exactly at the critical threshold.

Why the Point (0) Is Not the Problem

Many students initially think the issue is that (1/x) is undefined at (0).

However, measure theory ignores individual points because:

$$m({0})=0$$

A single point contributes no area.

In fact, the following theorem explains why.

Equality Almost Everywhere Theorem

If:

$$f=g\quad\text{almost everywhere}$$

then:

$$\int f,d\mu=\int g,d\mu$$

Changing a function on a set of measure zero does not affect its Lebesgue integral.

For example:

$$f(x)=0$$

and

have the same Lebesgue integral.

Therefore the failure of (1/x) to be integrable has nothing to do with the single point (0).

The Real Problem

The issue is not the value at (0).

The issue is the behavior near (0).

Every neighborhood around (0) contributes additional area.

For $(1/\sqrt{x})$, the accumulated area remains finite.

For (1/x), the accumulated area grows without bound.

The singularity is simply too strong.

Absolute Integrability

Another way to state the criterion is the following.

Absolute Integrability Theorem

If:

$$\int |f|,d\mu<\infty$$

then (f) is Lebesgue integrable.

If:

$$\int |f|,d\mu=\infty$$

then (f) is not Lebesgue integrable.

Applying this theorem to:

$$f(x)=\frac{1}{x}$$

gives:

$$\int_0^1 \left|\frac{1}{x}\right|,dx=\infty$$

and therefore the function is not integrable.

Geometric Interpretation

The distinction between ($1/\sqrt{x}$) and ($1/x$) is geometric.

Both functions become infinite near (0).

However:

• ($1/\sqrt{x}$) rises slowly enough that the total area remains finite.
• ($1/x$) rises rapidly enough that the accumulated area becomes infinite.

Thus the question is not:

Does the function become infinite?

The question is:

Does the total area remain finite?

Conclusion

The great insight of Lebesgue integration is that boundedness is not the decisive issue.

A function may be unbounded and still be perfectly integrable.

The true criterion is:

$$\int |f|,d\mu<\infty$$

The p-Integral Test explains precisely when singularities of the form (1/x^p) are integrable, while the Equality Almost Everywhere Theorem explains why individual points do not matter.

The function:

$$\frac{1}{\sqrt{x}}$$

is unbounded yet integrable.

The function:

$$\frac{1}{x}$$

is unbounded and not integrable.

The difference is not the existence of a singularity, but the amount of area that accumulates near it.

References

1. Terence Tao, An Introduction to Measure Theory, American Mathematical Society, 2011.
2. H. L. Royden and P. M. Fitzpatrick, Real Analysis, 4th Edition, Pearson, 2010.
3. Gerald B. Folland, Real Analysis: Modern Techniques and Their Applications, 2nd Edition, Wiley, 1999.
4. Walter Rudin, Real and Complex Analysis, 3rd Edition, McGraw-Hill, 1987.
5. Donald L. Cohn, Measure Theory, 2nd Edition, Birkhäuser, 2013.
6. Richard L. Wheeden and Antoni Zygmund, Measure and Integral: An Introduction to Real Analysis, CRC Press, 1977.