The Lebesgue Integral Review

# The Lebesgue Integral Review

We will now review some of the recent material regarding Lebesgue integrals.

- Recall from
**The Lebesgue Integral**page that a function $f$ is said to be**Lebesgue Integrable**on the interval $I$ if there exists upper functions $u$ and $v$ on $I$ such that:

\begin{align} \quad f = u - v \end{align}

- In other words, $f$ is Lebesgue integrable on $I$ if it is the difference of two upper functions on $I$. Furthermore, we define the
**Lebesgue Integral**of $f$ on $I$ to be the difference of the integrals of the upper functions, i.e.:

\begin{align} \quad \int_I f(x) \: dx = \int_I u(x) \: dx - \int_I v(x) \: dx \end{align}

- The set of all Lebesgue integrable functions on $I$ is denoted $L(I)$.

- We verified that the definition of the Lebesgue integral of a function is well-defined, that is, if $f = u_1 - v_1$ and $f = u_2 - v_2$ where $u_1, v_1, u_2, v_2 \in U(I)$ then:

\begin{align} \quad \int_I u_1(x) \: dx - \int_I v_1(x) \: dx = \int_I u_2(x) \: dx - \int_I v_2(x) \: dx \end{align}

- This guaranteed us that the value of the Lebesgue integral of $f$ on $I$ is independent on the choices of upper functions whose difference is $f$.

- On the
**Linearity of Lebesgue Integrals**page we saw the linearity properties hold for Lebesgue integrals. We saw that if $f, g \in L(I)$ then $(f + g) \in L(I)$ and:

\begin{align} \quad \int_I [f(x) + g(x)] \: dx = \int_I f(x) \: dx + \int_I g(x) \: dx \end{align}

- We also saw that if $f \in L(I)$ and for any $c \in \mathbb{R}$ we have that $cf \in L(I)$ and:

\begin{align} \quad \int_I cf(x) \: dx = c \int_I f(x) \: dx \end{align}

- We then looked at some nice comparison theorems for Lebesgue integrals on the
**Comparison Theorems for Lebesgue Integrals**. We saw that if $f \in L(I)$ and if $f(x) \geq 0$ almost everywhere on $I$ then:

\begin{align} \quad \int_I f(x) \: dx \geq 0 \end{align}

- We then proved that if $f, g \in L(I)$ and if $f(x) \leq g(x)$ almost everywhere on $I$ then:

\begin{align} \quad \int_I f(x) \: dx \leq \int_I g(x) \: dx \end{align}

- Furthermore, if $f, g \in L(I)$ and if $f(x) = g(x)$ almost everywhere on $I$ then:

\begin{align} \quad \int_I f(x) \: dx = \int_I g(x) \: dx \end{align}

- On the
**The Positive and Negative Parts of a Function**we defined two special functions. If $f$ is any function on $I$, then the**Positive Part of $f$**denoted $f^+$ is the nonnegative function defined for $x \in I$ by:

\begin{align} \quad f^+(x) = \max \{ f(x), 0 \} \end{align}

- Similarly, the
**Negative Part of $f$**denoted $f^{-}$ is the nonnegative function defined for $x \in I$ by:

\begin{align} \quad f^{-}(x) = \max \{ -f(x), 0 \} \end{align}

- Despite their names, both the positive and negative parts of $f$ are nonnegative functions. We then noted two very important identities regarding the positive and negative parts of $f$:

\begin{align} \quad f = f^+ - f^- \quad \mathrm{and} \quad \mid f \mid = f^+ + f^- \end{align}

- On the
**Lebesgue Integrability of the Positive and Negative Parts of a Function**page we saw that if $f \in L(I)$ then $f^+, f^- \in L(I)$.

- The result stated above was useful on the
**Lebesgue Integrability of the Absolute Value of a Function**page where we showed that if $f \in L(I)$ then $\mid f \mid \in L(I)$ and moreover:

\begin{align} \quad \biggr \lvert \int_I f(x) \: dx \biggr \rvert \leq \int_I \mid f(x) \mid \: dx \end{align}

- On the
**Additivity of Lebesgue Integrals on Subintervals of General Intervals**page we then looked at the Lebesgue integrability of functions on subintervals. We first saw that if $f \in L(I)$, and if $I_1, I_2 \subset I$ are subintervals such that $I_1$ and $I_2$ intersect in at most one point and such that $I = I_1 \cup I_2$ then $f \in L(I_1)$ and $f \in L(I_2)$ and furthermore:

\begin{align} \quad \int_I f(x) \: dx = \int_{I_1} f(x) \: dx + \int_{I_2} f(x) \: dx \end{align}

- We then proved that if $I_1, I_2 \subset I$ are subintervals such that $I_1$ and $I_2$ intersect in at most one point and such that $I = I_1 \cup I_2$, then if $g \in L(I_1)$, $h \in L(I_2)$, and $f$ is a function defined on all of $I$ by $f(x) = \left\{\begin{matrix} g(x) & x \in I_1\\ h(x) & x \in I_2 \setminus (I_1 \cap I_2) \end{matrix}\right.$ then $f \in L(I)$ and:

\begin{align} \quad \int_I f(x) \: dx = \int_{I_1} g(x) \: dx + \int_{I_2} h(x) \: dx \end{align}