# Continuity under Integral Sign

## Theorem

Let $\struct {X, \Sigma, \mu}$ be a measure space.

Let $U$ be a non-empty open set of a metric space.

Let $f: U \times X \to \R$ be a mapping satisfying:

$(1): \quad$ For all $\lambda \in U$, the mapping $x \mapsto \map f {\lambda, x}$ is $\mu$-integrable
$(2): \quad$ For $\mu$-almost all $x \in X$, the mapping $\lambda \mapsto \map f {\lambda, x}$ is continuous
$(3): \quad$ There exists a $\mu$-integrable $g: X \to \R$ such that:
$\forall \tuple {\lambda, x} \in U \times X: \size {\map f {\lambda, x} } \le \map g x$

Then the mapping $h: U \to \R$ defined by:

$\displaystyle \map h \lambda := \int \map f {\lambda, x} \map {\rd \mu} x$

is continuous.

## Proof

Let $\lambda_0 \in U$ be arbitrary.

Let $\sequence {\lambda_n}_{n \mathop \ge 1}$ be a sequence in $U$ which converges to $\lambda_0$.

Define the sequence of functions $f_n: X \to \R$, for $n = 0$ and $n \ge 1$ by $\map {f_n} x = \map f {\lambda_n, x}$.

By hypothesis $(1)$, for each $n \ge 1$, the function $f_n$ is $\mu$-integrable.

By hypothesis $(2)$, and by Sequential Continuity is Equivalent to Continuity in Metric Space, we have:

$\displaystyle \map f x = \lim_{n \mathop \to \infty} \map {f_n} x$

for $\mu$-almost all $x \in X$.

By hypothesis $(3)$, we have:

$\size {\map {f_n} x} \le \map g x$

for all $x \in X$ and all $n \ge 1$.

Therefore by Lebesgue's Dominated Convergence Theorem:

$\displaystyle \lim_{n \mathop \to \infty} \int f_n \rd \mu = \int f_0 \rd \mu$

That is:

$\displaystyle \lim_{n \mathop \to \infty} \int \map f {\lambda_n, x} \rd \mu = \int \map f {\lambda_0, x} \rd \mu$

or in terms of $h$:

$\displaystyle \lim_{n \mathop \to \infty} \map h {\lambda_n} = \map h {\lambda_0}$

Since the sequence $\sequence {\lambda_n}_{n \mathop \ge 1}$ was arbitrary, by Sequential Continuity is Equivalent to Continuity in Metric Space this shows that $h$ is continuous at $\lambda_0 \in U$.

But $\lambda_0 \in U$ was arbitrary, so $h$ is continuous in $U$.

$\blacksquare$