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The primitive of a continuous function on a compact interval is continuous via the Fundamental Theorem of Calculus. Let $I \subset \mathbb{R}$ be open and let $u': \overline{I} \mapsto \mathbb{R}$ be continuous. Then for any $y \in I$, by the $L^1-L^\infty$ bound from Holder's inequality, we have:

\begin{equation} |u(x)-u(y)|=\left|\int^x_y u'(z) \, dz \right| \leq \int^x_y \left| u'(z) \right| \, dz \leq \| u'\|_{L^{\infty}(\overline{I})} |x-y| \rightarrow 0 \quad \mbox{as } x \rightarrow y, \end{equation} so $u$ is continuous on $I$.

The same argument can be used if $u' \in L^p(I)$ if $p \in (1,\infty]$. However, I am unsure if it is still true if $u' \in L^1(I)$. Can I still conclude that $u$ is continuous?

Edit: fixing some errors in the question.

Edit 2: We use a different argument. The idea is to truncate $u'$ on smaller and smaller supports so that the resulting sequence $u_n'$ will tend to $0$ in the limit. The proof is concluded by the dominated convergence theorem (DCT) because $u_n' \leq u'$.

Fix $y \in I$. For each large positive integer $n$ (large enough of course so that $A_n \subset I$), let \begin{equation} A_n:=\{t \in I: y-1/n<t<y+1/n \}, \end{equation} and the indicator function \begin{equation} {\bf 1}_{A_n}(z):= \left\{ \begin{aligned} 1 \quad \mbox{if } z \in A_n,\\ 0 \quad \mbox{if } z \notin A_n.\\ \end{aligned} \right. \end{equation} Then $u'_n:={\bf 1}_{A_n}u'$ with $u'_n \rightarrow 0$ pointwise (a.e.) as $n \rightarrow \infty$ and $|u_n'| \leq u' \in L^1(I)$ by assumption. Hence, the DCT implies that \begin{equation} u(y+1/n)-u(y-1/n)\leq \int^{y+1/n}_{y-1/n} |u'(z)| \, dz = \int_I |u_n'(z)| \, dz \rightarrow \int_I 0 \, dz=0. \end{equation} That is, for each $\varepsilon > 0$, there exists an $n_0 > 0$ such that $n\geq n_0$ implies $\int^{y+1/n}_{y-1/n} |u'(z)| \, dz < \varepsilon$. Hence, for every $\varepsilon > 0$, there exists $\delta:=1/n_0$ such that \begin{equation} |x-y| < \delta=1/n_0 \quad \mbox{implies} \quad |u(x)-u(y)|\leq \int^{y+1/n}_{y-1/n} |u'(z)| \, dz<\varepsilon. \end{equation}

As Copperhat notes, this actually shows that $v << \mu$ where $v$ is a measure with density $u' \in L^1$ and $\mu$ is the Lebesgue measure.

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  • $\begingroup$ Precisely, a primitive of a continuous map on a compact interval is continuous on the interior of that interval. $\endgroup$
    – Yes
    Oct 31, 2015 at 5:26
  • $\begingroup$ Thanks i've added the clarification in the first sentence. $\endgroup$
    – Josh
    Oct 31, 2015 at 5:45
  • $\begingroup$ The subject line of this question says "Primitive of an $L^1$ function is continuous", without mentioning anything about derivatives. But the body of the question adds a complication: that the function you're integrating is a derivative. Everywhere? If the question is intended to be only about whether $x\mapsto\int_y^x f(z)\,dz$ is continuous when $f\in L^1,$ then the way it's expressed is too complicated. $\endgroup$ Feb 2, 2021 at 14:33

2 Answers 2

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Fact. If $\,f\in L^1[a,b]$, then for every $\varepsilon>0$, there exists a $\delta>0$, such that $I\subset [a,b]$ and $\mu(I)<\delta$ implies that $\int_I\lvert\, f\rvert \,dx<\varepsilon$.

Proof. If we set $E_M=\{x\in[a,b]: \lvert\, f(x)\rvert\ge M\}$, then $\lim_{M\to\infty}\int_{E_M}\lvert\, f\rvert\,dx=0$. To see this, let $$ f_M(x)=\left\{\begin{array}{ll} f(x) & \text{if} & \lvert\,f(x)\rvert\le M, \\ 0 & \text{otherwise.}\end{array}\right. $$ Then, the Monotone Convergence Theorem implies that $\lim_{M\to\infty} \int_{[a,b]}\lvert\, f_M\rvert \,dx=\int_{[a,b]}\lvert\, f\rvert \,dx$. But, $$ \int_{E_M}\lvert\, f\rvert\,dx= \int_{[a,b]}\lvert\, f\rvert \,dx-\int_{[a,b]}\lvert\, f_M\rvert \,dx. $$ Now let $\varepsilon>0$. First we obtain $M>0$, such that $\int_{E_M}\lvert\, f\rvert\,dx<\varepsilon/2$, and then we set $$ \delta=\frac{\varepsilon}{2M}. $$ Then, if $\mu(I)<\delta$, $$ \int_I\lvert\,f\rvert\,dx=\int_{I\cap E_M}\lvert\,f\rvert\,dx+ \int_{I\smallsetminus E_M}\lvert\,f\rvert\,dx\le \int_{E_M}\lvert\,f\rvert\,dx+ \int_{I}M\,dx < \frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon. $$

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The fundamental result you need is that if $f \in L^1[a,b]$, then the measure $\mu A = \int_A |f(x)| dx$ is absolutely continuous with respect to the Lebesgue measure. This means that for all $\epsilon>0$ there is some $\delta >0$ such that if $m A < \delta$ then $\mu A < \epsilon$.

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