# Continuity under Integral Sign

## Theorem

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

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

$\displaystyle h \left({\lambda}\right) := \int f \left({\lambda, x}\right) \, \mathrm d \mu \left({x}\right)$

is continuous.

## Proof

Let $\lambda_0 \in U$ be arbitrary.

Let $\left\langle{\lambda_n}\right\rangle_{n \ge 1}$ 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 $f_n \left({x}\right) = f \left({\lambda_n, x}\right)$.

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:

$\displaystyle f \left({x}\right) = \lim_{n \to \infty} f_n \left({x}\right)$

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

By hypothesis $(3)$, we have:

$\left|{f_n \left({x}\right)}\right| \le g \left({x}\right)$

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

Therefore by Lebesgue's Dominated Convergence Theorem:

$\displaystyle \lim_{n \to \infty} \int f_n \, \mathrm d \mu = \int f_0 \, \mathrm d \mu$

That is:

$\displaystyle \lim_{n \to \infty} \int f \left({\lambda_n, x}\right) \, \mathrm d \mu = \int f \left({\lambda_0, x}\right) \, \mathrm d \mu$

or in terms of $h$:

$\displaystyle \lim_{n \to \infty} h \left({\lambda_n}\right) = h \left({\lambda_0}\right)$

Since the sequence $\left\langle{\lambda_n}\right\rangle_{n \ge 1}$ 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 $U$.

$\blacksquare$