Hint for Lebesgue theory/functional analysis type of problem I am trying to solve the following problem, but I am not too familiar with functional analysis. Could you guys tell me where I should start? Thanks!
Let $f \in L^1(\mathbb{R})$ and define 
$$f_n(x) = \frac{1}{n} \int_x^{x+n} f(t)\,dt.$$
Show that $\|f_n\|_1 \leq \|f\|_1$ and $\|f_n-f\|_1 \to 0$ as $n \to 0$. 
This seem quite intuitive given $f_n$, but I have no idea where to start to formally prove it. Thank you so much!
 A: You are dealing with the convolution of $f$ with
$$
                     \phi_n(x) = \frac{1}{n}\chi_{[-n,0]}(x).
$$
That is,
$$
          (f\star\phi_n)(x)=\int_{-\infty}^{\infty}f(t)\phi_n(x-t)dt=\frac{1}{n}\int_{x}^{x+n}f(t)dt = f_n(x)
$$
Therefore $\|f_n\|_1 \le \|f\|_1\|\phi_n\|_1 = \|f\|_1$. If $g \in \mathcal{C}_c^{\infty}(\mathbb{R})$ (i.e., compactly supported, infinitely differentiable,) then
$$
\begin{align}
          \|f_n-f\|_1 & \le \|f\star\phi_n-g\star\phi_n\|_1+\|g\star\phi_n-g\|_1+\|g-f\|_1 \\
    & \le \|f-g\|_1+\|g\star\phi_n-g\|_1+\|g-f\|_1 \\
    & = 2\|f-g\|_1+\|g\star\phi_n-g\|_1
\end{align}
$$
First choose $g\in\mathcal{C}_c^{\infty}(\mathbb{R})$ so that $\|f-g\|_1 < \epsilon/3$ and then choose $n$ so that $\|g\star\phi_n-g\|_1 < \epsilon/3$.
A: For the first statement: as mentioned by Giovanni in the comments, you need to use Fubini's theorem, after a change of variables. After the change of variable $y=x-t$ in the inner integral, you have 
$$\|f_n\|_1= \int_{\mathbb R} \left| \frac1n \int_x^{x+n}f(t)\,dt\right|\, dx \leq \frac1n \int_{\mathbb R} \int_0^n |f(x-y)|\,dy\,dx,
$$
and now by Fubini you can swap the integrals and obtain the desired result (remember $L^1$ norm is translation invariant).
For the second, try to use the first part and dominated convergence theorem (you have an $L^1$ dominant and pointwise convergence) to obtain convergence in $L^1$ (which is what you are asked to prove).
