Criterion for a Measurable Function to be Lebesgue Integrable

Recall from the Measurable Functions page that a function $f$ is said to be measurable on $I$ if there sequences a sequence of step functions $(f_n(x))_{n=1}^{\infty}$ that converges almost everywhere to $f$ on $I$.

We said that the set of all measurable functions on an interval $I$ is denoted $M(I)$.

We have already established that all Lebesgue integrable functions are measurable functions, but the converse is not true. There exists functions that are measurable and not Lebesgue integrable. The following theorem gives us a criterion for when a measurable function is Lebesgue integrable.

Theorem 1: Let $f$ be a measurable function on an interval $I$ and suppose that there exists a Lebesgue integrable function $g$ on $I$ such that $\mid f(x) \mid \leq g(x)$ almost everywhere on $I$. Then $f$ is Lebesgue integrable on $I$.

• Proof: Let $f$ be a measurable function on the interval $I$ and suppose that there exists a Lebesgue integrable function $g$ on $I$ such that $\mid f(x) \mid \leq g(x)$ almost everywhere on $I$.

• Since $f$ is a measurable function on $I$ there exists a sequence of step functions $(f_n(x))_{n=1}^{\infty}$ that converges almost everywhere to $f$ on $I$, i.e. $\displaystyle{\lim_{n \to \infty} f_n(x) = f(x)}$ almost everywhere on $I$.

• For each $n \in \mathbb{N}$, define a new function $g_n$ on $I$ by:

(1)

\begin{align} \quad g_n(x) = \max \{ -g, \min \{ f_n, g \} \} \end{align}

• Let $(g_n(x))_{n=1}^{\infty}$ be the resulting sequence of these functions. Then each $g_n$ is Lebesgue integrable since we already know that the maximum/minimum of Lebesgue integrable functions is Lebesgue integrable.

• We further claim that $-g(x) \leq g_n(x) \leq g(x)$ i.e., $\mid g_n(x) \mid \leq g(x)$ almost everywhere on $I$ and for all $n \in \mathbb{N}$. To prove this claim, let $D \subseteq I$ be the set where $\mid f(x) \mid \not \leq g(x)$ if and only if $x \in D$. Then $m(D) = 0$ since $\mid f(x) \mid \leq g(x)$ almost everywhere on $I$. So, let $x \in I \setminus D$. Then either $f_n(x) = -g(x)$ or $f_n(x) = \min \{ f_n(x), g(x) \}$.

• If $g_n(x) = -g(x)$ then $-g(x) \leq g_n(x)$.

• If instead $g_n(x) = \min \{ f_n(x), g(x) \}$ then there are two subcases to consider. If $g_n(x) = f_n(x)$ then this implies that $-g(x) \leq f_n(x) \leq g(x)$. If $g_n(x) = g(x)$, then $g_n(x) \leq g(x)$. In all cases we see that:

(2)

\begin{align} \quad -g(x) \leq g_n(x) \leq g(x) \end{align}

• Furthermore, we claim that the sequence $(g_n(x))_{n=1}^{\infty}$ converges to $f$ almost everywhere on $I$. This is because since $(f_n(x))_{n=1}^{\infty}$ converges to $f(x)$ almost everywhere on $I$ we have that there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then $\mid f_n(x) \mid \leq g(x)$ almost everywhere on $I$. But then $g_n(X) = f_n(x)$, and so as $n \to \infty$, $g_n(x) \to f(x)$.

• So $(g_n(x))_{n=1}^{\infty}$ is a sequence of Lebesgue integrable functions that converges to $f$ almost everywhere on $I$ and is dominated by $g$. So, by Lebesgue's Dominated Convergence Theorem we have that the limit function $f$ is Lebesgue integrable on $I$. $\blacksquare$