# Does uniformly bounded sequence in Lp which converges almost everywhere converge in norm?

I'm trying to solve problem $$17$$ from chapter $$6$$ of Royden's Real Analysis. The problem is -

Let $$(f_n)_{n=1}^\infty$$ be a sequence of functions in $$L^p([0,1]), p\in(1,\infty)$$, which converge almost everywhere to some $$f\in L^p$$. Suppose that there is a constant $$M$$ s.t $$||f_n||_p\leq M$$ for all $$n$$. Then for each $$g\in L^q$$ (where $$\frac{1}{q}+\frac{1}{p}=1$$) we have -

$$\int fg=lim_{n\rightarrow \infty}\int f_ng$$

I want to solve this problem using the Riesz representation theorem for $$L^p$$ spaces, which states that any bounded linear functional in $$(L^p)^*=L^q$$ is of the form - $$\phi _g (h)=\int hg$$, where $$g\in L^q$$. Since $$g$$ corresponds to a bounded (and thus continuous) linear functional, the problem is basically solved, if I prove that $$f_n\rightarrow f$$ in norm. But I'm not sure that this is the case.

I know that generally, $$f_n\rightarrow f$$ almost everywhere does not imply convergence in norm. But since $$f_n$$ are uniformly bounded in norm by $$M$$, I'm hoping that I can somehow conclude that this is indeed the case (I know that if moreover $$||f_n||_p\rightarrow ||f||_p$$ then there actually is convergence in norm, maybe I can somehow use this fact?).

Does anyone know if this it true and how to prove this? If not, can anyone suggest a different approach?

It need not be the case that $$f_n \to f$$ in $$L^p$$.For example, consider $$f_n = n^{1/p} \chi_{[0,\frac{1}{n}]}$$ as functions on $$[0,1]$$ with the Lebesgue measure. Then $$f_n \to 0$$ a.e. but $$\|f_n\|_p = 1$$ so that $$f_n$$ is bounded in $$L^p([0,1])$$ but does not converge to $$0$$ in $$L^p([0,1])$$.
In fact, in Royden's book the Riesz representation theorem for $$L^p$$ is presented after this exercise so it's good to guess that we shouldn't use it.
Instead we will make use of Egoroff's theorem. First, since $$g \in L^q$$, for any $$\varepsilon > 0$$ there is a $$\delta > 0$$ such that $$m(E) < \delta$$ implies that $$\int_E |g|^q dm < \bigg(\frac{\varepsilon}{4M}\bigg)^q.$$
Then by Egoroff's theorem there is measurable $$E \subseteq [0,1]$$ such that $$m(E) < \delta$$ and $$f_n \to f$$ uniformly on $$F := [0,1] \setminus E$$. As a result, there is an $$N$$ such that $$n \geq N$$ and $$x \in F$$ implies that $$|f_n(x) - f(x)| \leq \frac{\varepsilon}{2m(F)^{1/p}\|g\|_q}.$$
Finally we get that for $$n \geq N$$ \begin{align} \bigg|\int_0^1 (f_n - f)g dm\bigg| \leq& \int_E |f_n-f||g| dm + \int_F |f_n - f||g|dm \\ \leq& \|f_n - f\|_p \frac{\varepsilon}{4M} + \bigg(\int_F |f_n - f|^pdm\bigg)^{1/p} \|g\|_q \\ \leq& \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon \end{align} where to get the second line I applied Holder's inequality and substituted the previous bound on the integral of $$|g|^q$$ over $$E$$. This is exactly the desired convergence result.