I need to prove the following property:
Let $f:\mathbb{R}^N\to \mathbb{R}$ a integrable function in $B(0,1)$. Then it is satisfied that $$\lim_{\varepsilon\to 0}\int_{|x|<\varepsilon}f(x)dx=0.$$
My attempt consists in trying to use the Dominated Convergence Theorem. I write $$\int_{|x|<\varepsilon}f(x)dx=\int_{\mathbb{R}^N}1_{B(0,\varepsilon)}f(x)dx.$$ For $\varepsilon<1$ we can bound $1_{B(0,\varepsilon)}f(x)$ by an integrable function, but I don't have a rigorous proof of the fact that $1_{B(0,\varepsilon)}f(x)\xrightarrow[]{\varepsilon\to0}0$ for each $x$. Sorry if this proof is obvious but I dont't see it in this moment. So, any help will be welcome.
This is very simple: suppose that $x_0\neq0$. Then $\|x_0\|>0$. For each $\varepsilon\in(0,\|x_0\|)$ it is $1_{\|x\|<\varepsilon}(x_0)=0$, so for any $x_0\neq0$ it is $$\lim_{\varepsilon\to0^+}1_{\|x\|<\varepsilon}(x_0)f(x_0)=0,$$ as this quantity is constantly $0$ eventually.
so $1_{\|y\|<\varepsilon}(x)f(x)\xrightarrow{\varepsilon\to0}0$ almost everywhere, since $\{0\}$ has measure $0$.
As you said, for any $\varepsilon>0$ it is $|1_{\|y\|<\varepsilon}(x)f(x)|\leq|f(x)|\in L^1$ (i.e. your functions are dominated), so you may apply DCT to get your result.