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 $\left( \lambda_n \right)_{n \geq 1}$ be a sequence in $U$ which converges to $\lambda_0$

Define the sequence of functions $f_n : X \to \R$, for $n = 0$ and $n \geq 1$ by $f_n(x) = f(\lambda_n, x)$.

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

By the hypothesis $(2)$, and by Sequential Continuity Equivalent to Continuity in Metric Space, we have
 * $\displaystyle f(x) = \lim_{n \to \infty} f_n(x)$

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

By the hypothesis $(3)$, we have $\left| f_n(x) \right| \leq g(x)$ for all $x \in X$ and all $n \geq 1$.

Therefore by Lebesgue's Dominated Convergence Theorem, we have
 * $\displaystyle \lim_{n \to \infty} \int f_n \, \mathrm d \mu = \int f_0 \, \mathrm d \mu$

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

Or in terms of $h$,
 * $\displaystyle \lim_{n \to \infty} h(\lambda_n) = h(\lambda_0)$

Since the sequence $\left( \lambda_n \right)_{n \geq 1}$ was arbitrary, by Sequential Continuity 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 $U$.