Convergence of particular function in $L^1$ I was struggling to prove or disprove a certain result. I was given that $\{Q_\alpha^K\}_{\alpha=1}^\infty$ are cubes of measure $1/K$ covering $\mathbb{R}^n$ for $K\in\{1,2,\ldots\}.$ I was also told $f\in L^1$ Then we define
$$f_K(x)=\frac{1}{|Q_\alpha^K|}\int_{Q_\alpha^K}f$$
The question I was asked was to prove or disprove that $f_K\to f$ in $L^1$. I know the result holds for continuous functions with compact support. In this case uniform continuity can be used to show $f_K\to f$ pointwise. Then Dominated Convergence can be used to show that $f_K\to f$ in $L^1$.
Loosing up the restriction to bounded with compact support, if our function was bounded and had compact support, then I believe the result could still work. My idea in that case was to use Lusin's Theorem. As we would be able to restrict all integrals over a compact set $C$. Lusin's Theorem I believe would allow use to essentially ignore the portion of $C$ were $f$ was not continuous, and the continuous part being handled by our previous case.
Now if the function didn't have compact support, or weren't bounded, could we apply similar logic by truncation? Or would we have to apply other techniques? Would the claim even work in that case? I will be pondering this question some more, but if anyone has any advice or a solution it would be much appreciated. Thank you.
 A: Yes it is true and your approach works. You need one or two additional things for the final result.
Firstly, it makes perfect sense to ask what happens regarding the convergence w.r.t $\| \cdot \|_{L^1}$ norm since:

*

*$$f_K \ \underset{K\to \infty}{\longrightarrow} \ f \ , \quad \text{a.e.}$$
and

*

*$$\| f_K \|_{L^1(\mathbb{R}^n)} \leq \| f \|_{L^1(\mathbb{R}^n)} < \infty \ .$$
Now bullet one is not that relevant to the question thus I will not prove it. Second bullet though is needed, so here's a sketch of the proof:
For any $K\in \mathbb{N}$, set where $A_K(x,y) = \frac{1}{|Q_{a,x}^{K}|} \chi_{Q_{a,x}^{K}}\otimes \chi_{Q_{a,x}^{K}}$, whenever $x\in Q_{a,x}^{K}$.  Then,

*

*$f_K(x) = \int_{\mathbb{R}^n} A_K(x,y) f(y) \ d\lambda_n(y)$

*$\int_{\mathbb{R}^n}A_K(x,y) \ d\lambda_n(y) \leq 1$

*$\int_{\mathbb{R}^n}A_K(x,y) \ d\lambda_n(x) \leq 1$
From Young's inequality for integral operators (see for example here), we have that, for any $K\in \mathbb{N}$,
$$\| f_K \|_{L^1(\mathbb{R}^n)} \leq \| f \|_{L^1(\mathbb{R}^n)} < \infty \ .$$
Now use the second bullet and the well known fact that continuous functions with compact support actually are $\| \cdot \|_{L^1}$-dense in $L^1(\mathbb{R}^n)$. Combined with your result about compactly supported functions , it basically over.
To be more precise, let $\epsilon > 0$. If $f\in L^1(\mathbb{R}^n)$, then there exists a $g\in C_c(\mathbb{R}^n)$, such that $\| f - g \|_{L^1(\mathbb{R}^n)}<\epsilon / 3$. Second bullet implies that $\| f_K - g_K \|_{L^1(\mathbb{R}^n)}<\epsilon / 3$, for all $K\in \mathbb{N}$, since $f_K - g_K = (f-g)_K$. Finally $\| g - g_K \|_{L^1(\mathbb{R}^n)} <\epsilon / 3$, for all large enough $K$, since $g\in C_c(\mathbb{R}^n)$. Thus, for all large enough $K$,
$$\| f - f_K \|_{L^1(\mathbb{R}^n)} \leq \| f - g \|_{L^1(\mathbb{R}^n)} + \| g - g_K \|_{L^1(\mathbb{R}^n)} + \| g_K - f_K \|_{L^1(\mathbb{R}^n)} < \epsilon$$
