Proof that $f$ is Lebesgue-integrable in $[0,1]$. Let $f:[0,1]\to\mathbb{R}$ be a non-negative function. For all $\epsilon\in(0,1]\,$, let $f$ be Riemann-integrable in $[\epsilon,1]$.
Show that $f\in L_{1}[0,1]\,$ iff $\,\,\lim_{\,\epsilon \to 0}\int_{\epsilon}^{1}f(x)\,dx\,$ exists. Moreover, in that case
$$\int_{[0,1]}f\,d\lambda=\lim_{\,\epsilon \to 0}\,\int_{\epsilon}^{1}f(x)\,dx$$
My attempt:
First, I want to show that $f$ is a measurable function. Since $f$ is Riemann-integrable in $[\epsilon,1]$, it is also Lebesgue-integrable and, in particular, it is measurable. Letting $\epsilon\to 0$, $f$ is a measurable function in $(0,1]$. Now, given any $t\in\mathbb{R}$, define the set $A$ by
$$A:= \{x\in[0,1] : f(x) \ge t  \}$$
We want to show that $A$ is a measurable set. Let $A_{1}:=A\,\cap\{0\}$ and $A_{2}:=A\cap (0,1]$. Since $f$ is measurable in $(0,1]$, $A_{2}$ is measurable. Now, the set $A_{1}$ will be empty or $A_{1}=\{0\}$, and in both cases is a measurable set. Hence, $A=A_{1}\cup A_{2}$ is measurable. So, $f$ is a measurable function in $[0,1]$.
Now, for all $n\ge 1$, let $E_{n}:=[1/n,1]$. Note that $E_{n} \subseteq E_{n+1}$ and $\bigcup E_{n}=(0,1]$. Further, since $f$ is non-negative, $(\int_{1/n}^{1} f(x)dx)_{n\ge 1}$ is an increasing sequence. So, it will converge iff it is bounded above. The claim then follows from
$$\lim_{\,\epsilon \to 0}\,\int_{\epsilon}^{1}f(x)\,dx=\lim_{\,n \to \infty}\,\int_{1/n}^{1}f(x)\,dx=\lim_{\,n \to \infty}\,\int_{E_{n}} f\,d\lambda=\int_{(0,1]} f\,d\lambda=\int_{[0,1]} f\,d\lambda< +\infty$$
Did I get this right?
 A: Your solution seem right but overcomplicated in the first part. Maybe a slightly more clear way to say the same is the following:
In general if $(f_n)$ is a sequence of (extended) real-valued measurable functions (for any chosen measure) and $f_n\to f$ pointwise almost everywhere (respect to this chosen measure) then $f$ is also measurable (see by example here).
So it follows that the function of your exercise is Lebesgue measurable in $[0,1]$, as $\mathbf{1}_{[\epsilon_n ,1]}f$ converges pointwise in $(0,1]$ to the same function for any chosen sequence $(\epsilon _n)\downarrow 0$ and the set $\{0\}$ have Lebesgue measure zero.
Also we have that $\int_\epsilon ^1f(x) dx=\int_{[0,1]}\mathbf{1}_{[\epsilon ,1]}f \mathop{}\!d \lambda $ for every chosen $\epsilon \in(0,1)$, therefore
$$
\lim_{\epsilon \to 0^+}\int_\epsilon ^1f(x) dx=\lim_{\epsilon \to 0^+}\int_{[0,1]}\mathbf{1}_{[\epsilon ,1]}f \mathop{}\!d \lambda \tag1
$$
Now as $\{\mathbf{1}_{[\epsilon ,1]}f :\epsilon \in(0,1)\}$ is a family of non-negative functions that increases to the same function as $\epsilon $ decreases, then the monotone convergence theorem give us
$$
\int_{[0,1]} f\mathop{}\!d \lambda =\lim_{\epsilon \to 0^+}\int_{[0,1]}\mathbf{1}_{[\epsilon ,1]}f \mathop{}\!d \lambda \tag2
$$
Then from $(1)$ and $(2)$ we have that
$$
\int_{[0,1]} f\mathop{}\!d \lambda =\lim_{\epsilon \to 0^+}\int_{\epsilon }^1f(x) \mathop{}\!d x\tag3
$$
So $f\in L^1$ if and only if the improper integral of Riemann $\lim_{\epsilon \to 0^+}\int_{\epsilon }^1f(x) \mathop{}\!d x$ converges in $\mathbb{R}$.
A: A more direct roof can be obtained by

*

*Recalling that
if $f$ is Riemann integrable over an interval $[a,b]$, then $f$ is Lebesgue measurable and integrable, and the integrals coincide.


*Applying dominated convergence $(|f_\varepsilon|\leq|f|$) and noticing that $\lim_{\varepsilon\rightarrow0}f_\varepsilon = f$ pointwise everywhere in $[0,1]$, except perhaps at $0$.
