Does integral of L1 function vanishes on small intervals

Question

I want to know if for $u \in L^1(\mathbb{R})$, we have $\int_0^\delta u \to 0 \text{ as } \delta \to 0$ ?

I know how to prove the result in $L^p(\Omega)$, $\Omega \subseteq \mathbb{R}^n$ and $p > 1$.

$ \left\vert \int_{B_q(\delta)} u \right\vert \leq \int_{B_q(\delta)} \vert u \vert$ where $B_q(\delta) \subseteq \Omega$ ($\delta$ small enough) is some ball in $q$-norm of radius $\delta$.

We have: $\int_{B_q(0, \delta)} \vert u \vert = \int_\Omega \vert u \vert \cdot 1_{B_q(0, \delta)}$, where $1_{B_q(0, \delta)}$ designates the indicator function on $B_q(0, \delta)$.

By Hölder inequality: $\int_\Omega \vert u \vert 1_{B_q(0, \delta)} \leq \Vert u \Vert_{L^p(\Omega)} \cdot \Vert 1_{B_q(0, \delta)} \Vert_{L^{p'}(\Omega)}$, with $p' < \infty$ the Hölder conjugate of $p$.

Since $\Vert 1_{B_q(0, \delta)} \Vert_{L^{p'}(\Omega)} \leq \Vert 1_{B_\infty(0, \delta)} \Vert_{L^{p'}} = \delta^n$, we get $\left \vert \int_{B_q(0, \delta)} u \right\vert \to 0$ as $\delta \to 0$.

For $p = 1$, we cannot use the Hölder inequality to prove the statement as $\Vert 1_{B_\infty(0, \delta)} \Vert_{L^\infty(\Omega)} = 1$.

Is the statement true for $p = 1$ and how to prove it?

EDIT:

Yes, the statement is true in $L^1$:

By a density theorem (e.g. Theorem IV.3 in [1]) for $\epsilon > 0$, there is some $\phi \in C(\Omega)$ such that $\Vert u - \phi \Vert_{L^1(\Omega)} \leq \epsilon$.

Then we have $\int_{B_q(\delta)} \vert u \vert \leq \int_{B_q(\delta)} \vert u - \phi \vert + \int_{B_q(\delta)} \vert \phi \vert$.

We have: $\int_{B_q(\delta)} \vert u - \phi \vert \leq \int_\Omega \vert u - \phi \vert \leq \epsilon$

And: $\int_{B_q(\delta)} \vert \phi \vert \leq \delta^n \Vert \phi \Vert_{L^\infty(B_\infty(\delta))}$

Since $\phi$ continuous in $\Omega$, then $\phi$ continuous in $\overline{B_\infty(\delta)}$ (the closure of the ball) and $\Vert \phi \Vert_{L^\infty(B_\infty(\delta))} < \infty$.

Then the right hand side of the inequality can be made as small as we want by playing on $\delta$, proving the result in $L^1$.

[1] Brézis, Haïm, Functional analysis. Theory and applications, Collection Mathématiques Appliquées pour la Maîtrise. Paris: Masson. 248 p. (1994). ZBL1147.46300.)

Answer 1

Yes, your solution is right. A more general result is this: let $(X,\mathcal A,\mu)$ a measure space and suppose that $\int_X |f|^pd\mu<\infty$ (for a measurable $f:X\to \Bbb R$ and some $p\in [1,\infty]$), then for every chosen $\epsilon>0$ there is a $\delta >0$ such that for any chosen $A\in\mathcal A$ it holds that $\mu(A)<\delta\implies\int_A |f|^pd\mu<\epsilon$.

Proof: the simple functions are dense in any $L^p$ space then there is a simple function $s$ such that $\|f-s\|_p<\epsilon/2$ for any chosen $\epsilon \in(0,1)$. Suppose that $p<\infty $, then as each simple function is bounded there is some $M\in(0,\infty )$ such that $|s|\leqslant M$, then by Minkowski's inequality we have that $$ \left(\int_{A}|f|^pd\mu\right)^{1/p}\leqslant \left(\int_{A}|f-s|^pd\mu\right)^{1/p}+\left(\int_A|s|^pd\mu\right)^{1/p}\leqslant \frac\epsilon 2+M(\mu(A))^{1/p} $$ Then choosing $\delta :=\left(\frac{\epsilon }{2M}\right)^p$ the statement is clear. For $p=\infty $ a shorter proof can be given.∎

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Alternatively to Minkowski's inequality we can use the inequality $$ |a+b|^p=2^p|\frac a2+\frac b2|^p\leqslant 2^p(\frac12|a|^p+\frac12| b|^p)=2^{p-1}(|a|^p+|b|^p) $$ for $p\in[1,\infty )$.