$\int _0^1 f(x)g_n (x) dx \rightarrow 0 $ as $n\rightarrow 0$

Question

Question:

Prove $\int _0^1 f(x)g_n (x) dx \rightarrow 0 $ as $n\rightarrow 0$ for $f\in \mathcal{L}^1([0,1])$ and $\{g_n\}_{n\in\mathbb{N}}$ a sequence of measurable functions on $[0,1]$ such that

(i) $|g_n(x)|\leq C$ a.e. $x\in [0,1]$ for some $C<\infty$, and

(ii) $\lim _{n\rightarrow\infty}\int _0^a g_n(x) dx=0$ for all $a\in [0,1]$.

My progress:

My first step is to show for $f = \phi$ a simple function, the above statement is true. Then by $f\in \mathcal{L}^1([0,1])$, there exists a simple function $\phi$ such that $\int _0^1 |f-\phi| dx < \epsilon$ for $\epsilon >0$. So $|\int _0^1 f(x)g_n (x) dx - \int _0^1 \phi(x)g_n (x) dx|\rightarrow 0 $ as $n\rightarrow 0$.

But there is a question that cannot be avoided, which is I always need to show that $\int _E |f||g_n| dx \rightarrow 0 $ for $E$ measurable and $m(E)<\epsilon$. For this problem, let $F = \{x\in [0,1]: |g_n(x)| > C \}$ and $\lim _{n\rightarrow\infty}m(F) = 0$. Now $\int _E |f||g_n| dx = \int _{E\cap F} |f||g_n| dx + \int _{E\cap F^c} |f||g_n| dx \leq \int _{E\cap F} |f||g_n| dx + \int _{E\cap F^c} C |f|dx $. $\int _{E\cap F^c} C |f|dx $ is less than $\int _{E} C |f|dx $. But for $\int _{E\cap F} |f||g_n| dx$, I am not sure how to formally prove it is not arbitrarily large.

If you have any comments or other way to solve this problem, please let me know. Thank you!

Answer 1

Hint:

It follows from the assumption on $g_n$ that $\int^b_a g_n\xrightarrow{n\rightarrow\infty}0$ for all $0\leq a<b\leq 1$ and so $\int^1_0 s g_n\xrightarrow{n\rightarrow\infty}0$ fo any step function $s$.

Now, step functions, i.e. functions if the for $\sum^N_{k=1}a_k\mathbb{1}_{(\alpha_k,\beta_k]}$, are dense in $L_1([0,1])$. Fix $f\in L_1[0,1]$, For any $\varepsilon>0$, there is a step function $s_\varepsilon$ on $[0,1]$ such that $\|f-s_\varepsilon\|_1<\varepsilon$. Then \begin{align} \Big|\int^1_0 fg_n\Big|&\leq\Big|\int^1_0(f-s_\varepsilon)g_n\Big| +\big|\int^1_0s_\varepsilon g_n\big|\\ &\leq C\|f-s_\varepsilon\|_1+\big|\int^1_0s_\varepsilon g_n\big| \end{align} Since $\lim_n\int^1_0s_\varepsilon g_n\xrightarrow{n\rightarrow\infty}0$, $$\limsup_n\Big|\int^1_0 fg_n\Big|\leq C\varepsilon$$

Answer 2

By part (ii), it follows that for any simple functions of the form $\phi = \sum_{i=1}^k \alpha_i \chi_{[0, a_i]}$ where $a_i \in [0,1]$ satisfy the following limit $$\lim_n \int_0^1 g_n \phi \, dx =0.$$ Let $\epsilon >0$ and $f \in L^1([0,1])$ be given. As every simple function defined on $[0,1]$ can be expressed as a linear combination of functions of the form $\chi_{[0,a]}$ where $a \in [0,1]$ and simple functions are dense in $L^1([0,1])$ there exists $\phi = \sum_{i=1}^k \alpha_i \chi_{[0, a_i]}$ so that $\int_0^1 |f - \phi| \, dx < \epsilon/C$. So, \begin{align*} \int_0^1 f g_n \, dx &= \int_0^1 (f - \phi) g_n \, dx + \int_0^1 \phi g_n \, dx\\ &\leq \int_0^1 |f - \phi| |g_n| \, dx + \int_0^1 \phi g_n \, dx\\ &\leq \epsilon + \int_0^1 \phi g_n \, dx \end{align*} Allowing $n \to \infty$ and as $\epsilon >0$ can be arbitrary small the result follows.