Continuity of Functions Defined by Lebesgue Integrals

# Continuity of Functions Defined by Lebesgue Integrals

Often times functions involving integrals are useful - especially in applied mathematics. We will now consider the various properties of functions defined by Lebesgue integrals. Let $I$ and $J$ be intervals and let $f : I \times J \to \mathbb{R}$ be a function. Then we can define a new, single variable function $F : J \to \mathbb{R}$ for all $y \in J$ by:

(1)\begin{align} \quad F(y) = \int_I f_y(x) \: dx = \int_I f(x, y) \: dx \end{align}

We will now give conditions for when the function $F$ is continuous on $J$.

Theorem 1: Let $I$ and $J$ be intervals in $\mathbb{R}$ and let $f : I \times J \to \mathbb{R}$. Suppose that:1) For every fixed $y \in J$, the function $f_y : I \to \mathbb{R}$ defined by $f_y(x) = f(x, y)$ is a measurable function on $I$.2) There exists a nonnegative function $g$ that is Lebesgue integrable on $I$ such that $\mid f_y(x) \mid \leq g(x)$ almost everywhere on $I$ and for all $y \in J$.3) For every fixed $y \in J$, $\displaystyle{\lim_{t \to y} f(x, t) = \lim_{t \to y} f_t(x) = f_y(x) = f(x, y)}$.Then we can conclude that: a) For every $y \in J$, $f_y$ is Lebesgue integrable on $I$, i.e., for every $y \in J$, $\displaystyle{\int_I f_y(x) \: dx = \int_I f(x, y) \: dx}$ exists.b) If $F : J \to \mathbb{R}$ is defined for all $y \in J$ by $\displaystyle{F(y) = \int_I f(x, y) \: dx}$ then $F$ is continuous on all of $J$, i.e., for every $y \in J$ we have that $\displaystyle{\lim_{t \to y} F(t) = \lim_{t \to y} \int_I f(x, t) \: dx = \int_I \lim_{t \to y} f(x, t) \: dx = \int_I f(x, y) \: dx = F(y)}$. |

**Proof of a):**Suppose that conditions (1), (2), and (3) all hold. For each fixed $y \in J$, since $f_y$ is a measurable function on $I$ and $\mid f_y(x) \mid \leq g(x)$ almost everywhere on $I$, where $g$ is a Lebesgue integrable function, then we have from the Criterion for a Measurable Function to be Lebesgue Integrable page that $f_y$ is Lebesgue integrable for every $y \in J$. In other words, the following Lebesgue integral exists for all $y \in J$:

\begin{align} \quad \int_I f_y(x) \: dx = \int_I f(x, y) \: dx \quad \blacksquare \end{align}

**Proof of b)**Now, fix $y \in J$ and let $(y_n)_{n=1}^{\infty}$ be any sequence of real numbers that converges to $y$, i.e., $\displaystyle{\lim_{n \to \infty} y_n = y}$. Now consider the sequence of functions $(f_{y_n})_{n=1}^{\infty}$. Each of these functions are Lebesgue integrable (as we proved above). Furthermore, $(f_{y_n})_{n=1}^{\infty}$ converges to $f_y$ almost everywhere on $I$ and additionally, the following inequality holds almost everywhere on $I$ and for all $n \in \mathbb{N}$:

\begin{align} \quad \mid f_{y_n}(x) \mid \leq g(x) \end{align}

- So by Lebesgue's Dominated Convergence Theorem we have that $f_y$ is Lebesgue integrable (though we already concluded that) and that:

\begin{align} \quad \int_I f_y(x) \: dx = \int_I \lim_{n \to \infty} f_{y_n}(x) \: dx = \lim_{n \to \infty} \int_I f_{y_n}(x) \: dx \quad (*) \end{align}

- Now, if $F$ is continuous at $y$ then from the Sequential Criterion for the Continuity of a Function we must have that $\lim_{n \to \infty} F(y_n) = F \left ( \lim_{n \to \infty} y_n \right ) = F(y)$. Evaluating the limit on the left and using $(*)$ and (3) gives us:

\begin{align} \quad \lim_{n \to \infty} F(y_n) = \lim_{n \to \infty} \int_I f_{y_n}(x) \: dx \overset{\mathrm{by} \: (*)} = \int_I \lim_{n \to \infty} f_{y_n}(x) \: dx = \int_I \lim_{n \to \infty} f(x, y_n) \: dx \overset{\mathrm{by} \: (3)} = \int_I f(x, y) \: dx = F(y) \end{align}

- So, by the sequential criterion for the continuity of a function, we must have that $F$ is continuous at each $y \in J$. $\blacksquare$