Probability with Measure

4.5 The Lebesgue integral

At last we are ready for the final step in the construction of the Lebesgue integral, which extends our current framework to a class of measurable functions that are real-valued, instead of just non-negative. This step requires a little preparation. We continue to work over an arbitrary measure space $(S, Σ, m)$ .

For a function $f : S \to R$ we define

$f_{+} = max (f, 0), f_{-} = max (- f, 0),$

which are pointwise definitions of functions $f_{+}, f_{-} : S \to R$ . It is easiest to see what is going on here with a picture:

Note that

$f = f_{+} - f_{-},$

and that by Theorem 3.1.5 both $f_{+}$ and $f_{-}$ are non-negative measurable functions from $S$ to $R$ . Step 2 (Definition 4.2.1) tells us the values of $\int_{S} f_{+} d m$ and $\int_{S} f_{-} d m$ , which are extended real numbers in $[0, \infty]$ . Note also that $| f | = f_{+} + f_{-}$ , and that $\int_{S} | f | d m$ is also covered by Step 2.

Definition 4.5.1 (Lebesgue Integral, Step 3) Let $f : S \to R$ be measurable. If at least one of $\int_{S} f_{+} d m$ and $\int_{S} f_{-} d m$ is not equal to $+ \infty$ , then we define

$\begin{matrix} (4.15) & \int_{S} f d m = \int_{S} f_{+} d m - \int_{S} f_{-} d m, \end{matrix}$

which is an extended real number.

If both $\int_{S} f_{+} d m$ and $\int_{S} f_{-} d m$ are equal to $+ \infty$ then $\int_{S} f d m$ is undefined. Note that in this case (4.15) would give $\infty - \infty$ , which is undefined.

It is important that we have a way to avoid handling infinities. This is provided by the following definition:

For non-negative measurable functions part 1 is contained within Theorem 4.4.1. Thus

\int_{A} f_{\pm} d m + \int_{B} f_{\pm} d m = \int_{A \cup B} f_{\pm} d m

. Subtracting the

f_{-}

case from the

f_{+}

case gives

$\int_{A} f_{+} d m - \int_{A} f_{-} d m + \int_{B} f_{+} d m - \int_{B} f_{-} d m = \int_{A \cup B} f_{+} d m - \int_{A \cup B} f_{-} d m$

which by (4.15) is exactly what we need to prove part 1.

For parts 2-5, it suffices to prove the case $S = A$ . Note that in view of Definition 4.1.3 we can recover the general case by replacing $f$ and $g$ by $𝟙_{A} f$ and $𝟙_{A} g$ . Part 5 follows immediately from Lemma 4.2.6 and (4.15). Several of the remaining parts are left for you as excerises. Part 3 is Exercise 4.4 part (c). Part 4 is Exercise 4.4 part (a). For part 2, Exercise 4.4 part (d) shows that

$\begin{matrix} (4.17) & α \int_{S} f d m = \int_{S} α f d m \end{matrix}$

for all $f \in L^{1}$ . In order to prove part 2 it remains only to show that

$\begin{matrix} (4.18) & \int_{S} f + g d m = \int_{S} f d m + \int_{S} g d m . \end{matrix}$

We will prove (4.18) here. Note that we may assume that both $f, g$ are not identically $0$ , because $\int_{S} 0 d m = 0$ . The fact that $f + g$ is in $L^{1}$ if $f$ and $g$ are follows from Exercise 4.4 part (b).

To show (4.18) we first need to consider six different special cases. Writing $h = f + g$ , these cases are (1) $f \geq 0, g \geq 0, h \geq 0$ , (2) $f \leq 0, g \leq 0, h \leq 0$ , (3) $f \geq 0, g \leq 0, h \geq 0$ , (4) $f \leq 0, g \geq 0, h \geq 0$ , (5) $f \geq 0, g \leq 0, h \leq 0$ , (6) $f \leq 0, g \geq 0, h \leq 0$ . Note that case (1) is precisely Lemma 4.3.2. We’ll just prove case (3) here, and note that the others are similar. To show case (3) we write $f = h + (- g)$ and, noting that all these are non-negative functions, from Lemma 4.3.2 we obtain $\int_{S} f d m = \int_{S} (f + g) d m + \int_{S} (- g) d m$ , and hence from (4.17) we have $\int_{S} (f + g) d m = \int_{S} f d m - \int_{S} (- g) d m = \int_{S} f d m + \int_{S} g d m,$ which proves case (3).

For a general $f \in L^{1}$ , we write $S = S_{1} \cup S_{2} \cup S_{3} \cup S_{4} \cup S_{5} \cup S_{6}$ , where $S_{l}$ is the set of all $x \in S$ for which case ( $l$ ) holds for $l = 1, 2, \dots, 6$ . These sets are disjoint and measurable. We then have

$\int_{S} (f + g) d m = \sum_{i = 1}^{6} \int_{S_{i}} (f + g) d m = \sum_{i = 1}^{6} \int_{S_{i}} f d m + \sum_{i = 1}^{6} \int_{S_{i}} g d m = \int_{S} f d m + \int_{S} g d m .$

In the above, the first and last equalities use part 1 of the present lemma, and the middle equality uses cases (1)-(6) above. This completes the proof of part 2. ∎