Lebesgue integral · Definition
Measurable function
Measurability lets us define Lebesgue integrals without partitioning the range. It is a necessary requirement: without it the Lebesgue integral is not defined.
Definition
A function f: X → ℝ is measurable with respect to a σ-algebra \(\mathcal{A}\) on X if the preimage \(f^{-1}(B)\) belongs to \(\mathcal{A}\) for every Borel set B of ℝ.
It suffices to check the condition on a generating family of Borel sets, such as intervals \((a,\infty)\), and continuous functions are always measurable with respect to the Borel σ-algebra. Sums, products and pointwise limits of measurable functions remain measurable.
Let \(f(x)=x^2\) on \(\mathbb{R}\) with the Borel σ-algebra. Since \(\{x: x^2>a\}\) is a union of open intervals for each \(a\), checking that family is enough to conclude that f is measurable.
A typical mistake is checking only a few specific Borel sets and concluding measurability; one must use a generator of the σ-algebra. Another error is confusing measurability with continuity, which is only a sufficient condition.
Think about it
Why is checking that \(f^{-1}((a,\infty))\) is measurable for a single \(a\) not enough to conclude that f is measurable?
Because the definition requires \(f^{-1}(B)\in\mathcal{A}\) for every Borel B, and one interval does not generate the whole Borel σ-algebra, so a single isolated check does not determine measurability.
Exercises 10
- What is the name of the condition satisfied by f when \(f^{-1}(B)\) lies in \(\mathcal{A}\) for every Borel B?
- Which σ-algebra is usually used on ℝ to define measurability of real functions?
- Which of these functions from ℝ to ℝ is NOT measurable with respect to the Borel σ-algebra?
- What kind of set is the preimage \(f^{-1}((a,\infty))\) when f is measurable?
- Is every continuous function from ℝ to ℝ measurable with respect to the Borel σ-algebra?
- Let \(f(x)=\begin{cases}1 & x\ge 0\\ 0 & x<0\end{cases}\) on ℝ. What is \(f^{-1}((1/2,\infty))\)?
- Let \(f_n(x)=x^n\) on \([0,1]\). What is the pointwise limit \(f\) and why is it measurable?
- If f and g are measurable, why is \(f+g\) measurable?
- Let \(f\) be measurable with \(f^{-1}((a,\infty))\in\mathcal{A}\) for all \(a\in\mathbb{R}\). What does this imply about \(f^{-1}([a,\infty))\)?
- Let \(f\) be measurable and \(g(x)=f(x)^2\). Prove that \(g\) is measurable, stating which set you check for \(a<0\) and for \(a\ge 0\). What is the correct conclusion?
References
- Gerald B. Folland, Real Analysis: Modern Techniques and Their Applications, 2nd edition, Wiley, 1999.
- H. L. Royden, P. M. Fitzpatrick, Real Analysis, 4th edition, Pearson, 2010.