Uniform convergence of integral when integrand converges pointwise

Let $g(x)$ be a continuous real valued function defined on $[-a,1+a]$ where $a>0.$

Let $f(x)$ be a continuous real valued function, $f\left( x\right) \geq 0$ for $x\in \lbrack 0,1],$ equal to zero otherwise.

Let $g_{n}(x)$ be a continuous real valued function with domain $% [-a,1+a].$

Let $g_{n}\left( x\right) \rightarrow g\left( x\right)$ for any $x\in \lbrack -a,1+a],$ so that $g_{n}$ converges to $g$ pointwise.

For $\left\vert \Delta \right\vert <a,$ define

$B_{n}(\Delta )=\int_{0}^{1}g_{n}(x+\Delta )f(x)dx$

$B(\Delta )=\int_{0}^{1}g(x+\Delta )f(x)dx$

Since $g_{n}(x+\Delta )$ converges pointwise to $g(x+\Delta ),$ $% B_{n}(\Delta )\rightarrow B(\Delta )$.

Question: under what conditions is the convergence uniform in $\Delta ?$

That is, can we find conditions such that for any $\varepsilon >0,\exists N$ such that $n>N\Rightarrow \left\vert B_{n}(\Delta )-B(\Delta )\right\vert <\varepsilon \forall \Delta \in [-a,a]$ ?

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Under the given conditions, $B_n$ does not necessarily converge pointwise to $B$; you need a dominating function for the $g_n$, or some other condition ensuring uniform integrability. –  Nate Eldredge Oct 6 '12 at 19:08

It is sufficient that $g_n \to g$ uniformly.