Show that $g(x) = \int_E f(x − t)d\lambda(t)$ is continuous on $\mathbb{R}$ Let f be a function in $L^1 = L^1(\mathbb{R}, \lambda)$, where  $\lambda$ is the Lebesgue measure on $\mathbb{R}$. Let $E$ be a measurable set with $\lambda(E) < \infty$.
Prove that the following function is continuous on $\mathbb{R}$
$$g(x) = \int_E f(x − t)d\lambda(t)$$
You may use without proof that "for each $\epsilon > 0$ there exists a continuous
function $\phi$ on R with compact support such that $\|f − \phi\|_1 <\epsilon$ ".
Is the following proof working?
fix $\delta>0$, and let the compact set to be such that $A = \{x \in E : |x-y| \leq \delta ,\forall y \in E\}$ so $\lambda(A)\leq \delta$. I have to show that $\|g(x)-g(y)\|<\epsilon$. 
By above lemma there exists continuous function $\phi$ such that $\|f(x-t) − \phi(x)\|_{L^1(A)} <\frac{\epsilon}{3}$ and $\|f(y-t) − \phi(y)\|_{L^1(A)} <\frac{\epsilon}{3}$ , also let by continuity of $\phi$ let $\|\phi(x)-\phi (y)\|_{L^1(A)}\leq \frac{\epsilon}{3}$
\begin{align}
\|g(x)-g(y)\|_{L^1(A)} 
& = |\int_Af(x-t)-\phi(x)+\phi(x)-\phi(y)+\phi(y)-f(y-t)d\lambda(t)|\\ 
& \leq \int_A |f(x-t)-\phi(x)|d\lambda(t)+\int_A |\phi(x)-\phi(y)|d\lambda(t)+\int_A |\phi(y)-f(y-t)|d\lambda(t)\\
&  = \|f(x-t) − \phi(x)\|_{L^1(A)} + \|\phi(x) − \phi(y)\|_{L^1(A)} +\|f(y-t) − \phi(y)\|_{L^1(A)} \\
& \leq \frac{\epsilon}{3} +\frac{\epsilon}{3}+\frac{\epsilon}{3}\\ 
& = \epsilon
\end{align}
 A: There is a considerable amount of confusion going on here. The fact you are given says that given $\epsilon > 0$, there exists a compactly supported continuous function $\phi$ such that
$$\int_{\mathbb{R}} \vert f(t) - \phi(t) \vert d\lambda(t) < \epsilon.$$
Notice that in your integrals in the last few lines, $\phi(x)$ does not depend on the variable $t$ with respect to which you are integrating, so the estimates are not valid.
Secondly, you have chosen some set $A$ to integrate over. This doesn't make sense. You are trying to estimate $\vert g(x) - g(y) \vert$, where $g$ is a function taking a parameter $x$ to an integral over $E$ of a function depending on $x$. You are looking at the $L^1$ norm on $A$ of $g(x) - g(y)$, which is not the same as changing domain of integration in the integral defining $g$ to $E$.
There is also a major problem right at the start of your proof: for each $x \in \mathbb{R}$ you have a function $t \to f(x - t)$. There is no guarantee that all of these functions can be simultaneously approximated in $L^1$ by a single continuous, compactly supported $\phi$. What we can do is approximate the single function $f$, i.e., we can choose a continuous function $\phi$ with compact support $\phi$ such that $\Vert f - \phi \Vert_{L^1(\mathbb{R})} < \frac{\epsilon}{3}$.
For the sake of brevity let
$$g_h(x) = \int_Eh(x - t)d\lambda(t)$$
for a function $h \in L^1(\mathbb{R})$. In particular $g = g_f$. Then for $x,y \in \mathbb{R}$ we have
\begin{equation*}
\begin{aligned}
g(x) - g(y) &= g_f(x) - g_{\phi}(x) + g_{\phi}(x) - g_{\phi}(y) + g_{\phi}(y) - g_f(y),
\end{aligned}
\end{equation*}
thus
$$\vert g(x) - g(y) \vert \leq \vert g_f(x) - g_{\phi}(x) \vert + \vert g_{\phi}(x) - g_{\phi}(y) \vert + \vert g_{\phi}(y) - g_f(y) \vert,$$
so the goal is to show that the three terms above can be made $< \frac{\epsilon}{3}$ by choosing $x,y$ close enough.
The first term is
$$\left\vert \int_E (f(x - t) - \phi(x - t)) d\lambda(t)\right\vert.$$
This is bounded by $\Vert f - \phi \Vert_{L^1(\mathbb{R})}$, which is $< \frac{\epsilon}{3}$ by our choice of $\phi$.
The third term is essentially identical. The middle term is
$$\left\vert \int_E (\phi(x - t) - \phi(y - t)) d\lambda(t) \right\vert.$$
Here you need to use the following fact: since $\phi$ is continuous on $\mathbb{R}$ and compactly supported, in fact $\phi$ is uniformly continuous on $\mathbb{R}$. Since $E$ has finite measure, you can use uniform continuity to choose $\delta > 0$ such that if $\vert x - y \vert < \delta$, then the above is $< \frac{\epsilon}{3}$. Note that we don't have to fix any particular $x$ or $y$ here before choosing $\delta$ ($g$ is actually uniformly continuous).
The point of choosing a continuous compactly supported function really is to get a uniformly continuous, integrable function. But then we have the $x$ and $y$ to deal with as well as the $f$ and $\phi$. What we do is "change one at at a time" (first change $x$ to $y$ and then change $f$ to $\phi$ using the triangle inequality, leaving us with three quantities all of which we can estimate). This is a common technique throughout analysis: you want to estimate how some function $A(x)$ changes when the parameter $x$ is changed slightly. It would be much easier if $A$ had some nice properties, but it turns out $A$ can be approximated by some other function $B(x)$ which does have these nice properties. Then we use the triangle inequality:
$$\vert A(x) - A(y) \vert \leq \vert A(x) - B(x) \vert + \vert B(x) - B(y) \vert + \vert B(y) - A(y) \vert.$$
