Your assumption "Show that if $f$ is Lebesgue-integrable" amounts to assuming that
$$
\int_{(0,1]} |f| < \infty.
$$
Suppose first that $f$ is positive. Then the function
$$
g(x)= \int_{[x,1]} f
$$
(with the integral being the Lebesgue integral) is a continuous function of $x$ and is bounded above by assumption. Since $g(x)$ increases as $x$ decreases towards $0$, the limit
$$
\lim_{x \searrow 0^+} g(x) = \int_{(0,1]}f
$$
exists and is equal to the Lebesgue-integral of $f$, because
$$
\int_{(0,x)} f \longrightarrow 0
$$
is the difference between $g(x)$ and the integral of $f$ over $(0,1]$ and it goes to zero. Since the Riemann integral over the interval $[x,1]$ and the Lebesgue integral over the same interval coincide, the improper integral exists in this case.

To complete the proof, write
$$
f_+ = \max \{f,0\}, \qquad f_- = \max\{-f,0\}.
$$
Notice that $f = f_+ - f_-$, and since both functions are positive and continuous you can apply the result above. By the linearity of both integrals (Riemann and Lebesgue) over sums of continuous functions, you are done.

Hope that helps,

You mean to be Lebesgue integrable and not Riemann integrable? The answer is yes. Classic example, let $f(x)=1$ if $x$ is a rational number and zero otherwise on the interval [0,1].

By the way, the Lebesgue integral is a generalization of the Riemann integral. Every Riemann integrable function is Lebesgue integrable. On the other hand there are plenty of functions which are Lebesgue integrable but not Riemann integrable.

Addendunm:

To the OP, looking at the comment below, I think there are several confusions here. You seem to be confusing an antiderivative with the integral. Antiderivatives and integrals are two entirely different things.

Sticking with Riemann integral, a Riemann integral is defined as the (signed) area under a curve so the Riemann integral of a function is always a number. Your textbook calls this the definite integral.

An antiderivative of $f(x)$ is a function is another function $F(x)$ such that $F'(x)=f(x)$. An antiderivative is always a function.

They are two different things but Newton and Leibniz proved that they are actually very closely related (a very useful thing by the way). The fundamental theorem of calculus says that in order to find the (definite) integral of a function, just compute its antiderivative and evaluate it at the top minus the bottom, meaning

$$\int_a^b f(x)dx=F(b)-F(a).$$

So when we say that a function is integrable, we mean that we can find the area underneath it. If you mean antiderivative then say antiderivative. Don't call the antiderivative the integral or the other way around.

Now to answer what I think you originally asked, Lebesgue integral doesn't use antiderivatives. Only the Riemann integral does. They both measure area under the curve but they use different methods. Whenever they both exist, they both agree and give you the same number but they use different methods so it doesn't even make sense to ask "if a function can be a Lebesgue integral". There is no such thing as "an indefinite Lebesgue integral". The Lebesgue integral is useful because it works on many functions that the Riemann integral can't handle BUT the Riemann integral is much easier to do/understand/develop and it also came first historically. So the Lebesgue integral is a (great) generalization of the Riemann integral so the classic example I gave you is one such function which Lebesgue can handle but Riemann can't.

Second Addendum:

Here's Martin's answer from the comments below

"In Lebesgue theory the primitive $F(x)=\int_a^x f(t)dt$ makes sense for every integrable function $f:[a,b]\rightarrow\mathbb{R}$ and by the Lebesgue differentiation theorem a function $F$ is a primitive iff it is absolutely continuous. Perhaps OP asks whether the derivative of an absolutely continuous function is always Riemann integrable. In other words, whether there is an integrable function that is not a.e. equal to a Riemann integrable function (yes: characteristic function of a fat Cantor set)."

## Best Answer

What you write is more or less true. First note that it is not enough to assume that $f(x,t)$ is differentiable

at $x_0$(for (almost) all $t$), we need that there is a neighorhood $(x_0 - \varepsilon, x_0 + \varepsilon) =:I$ such that $f(\cdot, t)$ is differentiable in $x$ for all $x \in I$ and (almost) all $t \in A$, where the "exceptional null set" mustnotdepend on $x$.The assumption $|\partial f/\partial x (x,t)| \leq g(x)$ only makes sense in this setting, otherwise we could only require $x_0$ instead of $x$ and the proof also needs this assumption (see below).

Now, let $h_n \to 0$ with $h_n \neq 0$ for all $n$ and w.l.o.g. with $x_0 + h_n \in I$ for all $n$. Let $$ F(x) := \int_A f(x,t) \, dt \text{ for } x \in I,\\ G(x) := \int_A \frac{\partial f}{\partial x}(x,t) \, dt \text{ for } x \in I. $$ Here, the integral defining $F$ is interpreted as an improper Riemann integral (it is possible that this integral does not exist as a Lebesgue integral), but the integral defining $G$ is interpreted as a Lebesgue integral. If $t \mapsto \frac{\partial f}{\partial x}$ is Riemann integrable over every compact subinterval of $A$ (for example, if it is continuous in $t$), then it can also be interpreted as an improper Riemann integral, at least if we assume that the dominating function $g$ is (improper) Riemann integrable.

Observe that the assumptions guarantee that $F,G$ are well-defined. Measurability of $t \mapsto \frac{\partial f}{\partial x}(x,t)$ for each $x \in I$ is implied by the pointwise convergence $$\frac{\partial f}{\partial x}(x,t) = \lim_n \frac{f(x+h_n, t) - f(x,t)}{h_n}, $$ which will also be employed below.

We now have \begin{eqnarray*} \left|\frac{F\left(x_{0}+h_{n}\right)-F\left(x_{0}\right)}{h_{n}}-G\left(x_{0}\right)\right| & = & \left|\int_{A}\frac{f\left(x_{0}+h_{n},t\right)-f\left(x_{0},t\right)}{h_{n}}-\frac{\partial f}{\partial x}\left(x_{0},t\right)\,{\rm d}t\right|\\ & \leq & \int_{A}\left|\frac{f\left(x_{0}+h_{n},t\right)-f\left(x_{0},t\right)}{h_{n}}-\frac{\partial f}{\partial x}\left(x_{0},t\right)\right|\,{\rm d}t. \end{eqnarray*} Note that the integrands are measurable and that for (almost) all $t \in A$, the mean value theorem yields some $\xi_{n,t} \in I$ (even between $x_0$ and $x_0 +h_n$) such that $$ \left|\frac{f\left(x_{0}+h_{n},t\right)-f\left(x_{0},t\right)}{h_{n}}\right|=\left|\frac{\partial f}{\partial x}\left(\xi_{n,t},t\right)\right|\leq g\left(t\right). $$ This shows that the integrands in the integrals are Lebesgue-integrable and dominated by an integrable function. As the integrand in the last integral converges pointwise to $0$ for $n \to \infty$, the dominated convergence theorem yields $$ \left|\frac{F\left(x_{0}+h_{n}\right)-F\left(x_{0}\right)}{h_{n}}-G\left(x_{0}\right)\right| \to 0, $$ so that $F$ is differentiable (in $x_0$) with the expected derivative.