Continuity of Functions Defined by Lebesgue Integrals

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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

\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 $y \in J$ :

(2)

\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 $n \in \mathbb{N}$ :

(3)

\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:

(4)

\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:

(5)

\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 $y \in J$ . $\blacksquare$

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Continuity of Functions Defined by Lebesgue Integrals