Continuity of the lebesgue integral How does one show that the function, $g(t) = \int \chi_{A+t} f $ is continuous, given that $A$ is measurable, $f$ is integrable and $A+t = \{x+t: x \in A\}$.
Any help would be appreciated, thanks
 A: The result is not true in general, for example, let $\mu A = 1_A(0)$ on $\mathbb{R}$, $f(x) = 1$ and $A = \{0\}$. Then $g(t) = 1_{\{0\}} (t)$ which is not continuous.
The result is true for the Lebesgue measure. Let $f_t$ denote the function $f_t(x) = f(x-t)$. Then we have $g(t) = \int 1_{A+t} f = \int 1_A f_t = \int_A f_t$. Since $g(s+t) = \int_A (f_s)_t$, we see that it is sufficient to show that $g$ is continuous at $0$ (that is, assuming that the result is true for integrable $f$).
Littlewood's principles tells us that $C_0(\mathbb{R})$ is dense in $L^p(\mathbb{R})$, so for $\epsilon>0$, we can find a $\tilde{f} \in C_0(\mathbb{R})$ such that $\|f-\tilde{f}\|_1 < \frac{\epsilon}{3}$. Since $\tilde{f} \in C_0(\mathbb{R})$, it is uniformly continuous and supported on a set of finite measure, hence we can find a $\delta>0$ such that if $|t| <  \delta$, then $\|\tilde{f} - \tilde{f}_t\| < \frac{\epsilon}{3}$. Consequently we have
$$\|f-f_t\| \leq \|f -\tilde{f}\|+ \|\tilde{f}-\tilde{f}_t\|+ \|\tilde{f}_t - f_t\| < \epsilon$$
So, if $|t| < \delta$, then $|g(t)-g(0)| \leq \int_A |f-f_t|\leq \|f-f_t\| < \epsilon$.
A: Notice that 
$$
|g(t+h)-g(t)| \le \int_{(A+t)\Delta A} |f|
$$
so it is enough to prove that
$$
  |(A+t)\Delta A| \to 0 \qquad \text{as }t \to 0
$$
where $\Delta$ is the symmetric difference, since 
$$
\int_{A_k} f \to 0 
$$
if $|A_k|\to 0$.
A: Assuming the integral is using Lebesgue measure over ${\Bbb R}$, one approach is to use the fact that continuous functions with compact support are dense in $L^1$.  Then, observe that if $f$ is continuous with compact support, $x\mapsto f(x+h)$ converges uniformly to $f$ as $h\to 0$.
