Lebesgue integral, differentiation under the integral sign

Question

Let $f:[0,1] \times [0,1] \to \mathbb R$ be a function such that:

(a) for each $x \in [0,1]$, the function $y \to f(x,y)$ is Lebesgue integrable on $[0,1]$.

(b) $\dfrac{\partial f}{\partial x}(x,y)$ is a bounded function of $(x,y)$.

Show that for each $x$, the function $y \to \dfrac{\partial f}{\partial x}(x,y)$ is measurable and $\dfrac{d}{dx} \int_0^1 f(x,y)dy=\int_0^1 \dfrac{\partial f}{\partial x}(x,y)dy$.

I am a bit stuck on the exercise, to prove that $g_x(y)=\dfrac{\partial f}{\partial x}(x,y)$ I did the following:

If for each $x$, $f_x(y)=f(x,y)$ is integrable, by definition, it is measurable. So given $x$ in [0,1), the function $f_{x+\frac{1}{n}}$ is measurable for each $n>\dfrac{1}{x+1}$ and the function $f_{1-\frac{1}{n}}$ is measurable for $x=1$, but then for $0 \leq x <1$ we have $$\dfrac{\partial f}{\partial x}(x,y)=\lim_{n \to \infty} \dfrac{f(x+\frac{1}{n},y)-f(x,y)}{\frac{1}{n}}$$

and for $x=1$, $$\dfrac{\partial f}{\partial x}(x,y)=\lim_{n \to \infty} \dfrac{f(x-\frac{1}{n},y)-f(x,y)}{-\frac{1}{n}}$$

Since each function is a pointwise limit of measurable functions, then for each $0 \leq x \leq 1$, $\dfrac{\partial f}{\partial x}(x,y)$ is a measurable function.

I am not sure if my reasoning is correct and I don't know what to do for the last part of the exercise. Any help would be greatly appreciated. Thanks in advance

Answer 1

The measurability of $y\to\frac{\partial f(x,y)}{\partial x}$ is OK. For the differentiation under the integral sign - I follow Theorem 2.75 here.

Take an arbitrary $x_0\in (0,1)$ and denote $\Phi(x)=\int\limits_{0}^{1}{f(x,y)dy}\,$. Let $\{x_n\}_{n=1}^{\infty}\subset [0,1]$ is an arbitrary sequence tending to $x_0$: $$\left(\frac{d}{dx}\int\limits_{0}^{1}{f(x,y)dy}\right)_{|x=x_0}=\frac{d\Phi}{d x}(x_0)=\lim\limits_{n\to\infty}{\frac{\Phi(x_n)-\Phi(x_0)}{x_n-x_0}}=\\ \lim\limits_{n\to\infty}{\int\limits_{0}^{1}{\underbrace{\frac{f(x_n,y)-f(x_0,y)}{x_n-x_0}}_{g_n(y)\to\frac{\partial f}{\partial x}(x_0,y)\,\,a.e\,\,y\in [0,1]}dy}}$$ So we have a pointwise a.e (in $y$) convergence of the sequence of functions $g_n(y)$ under the last integral to the function $\frac{\partial f}{\partial x}(x_0,y)$. Now to apply Lebesgue Dominated Convergence theorem we only need this sequence to be bounded by some summable function in $y$. We use the MVT: $$\left | \frac{f(x_n,y)-f(x_0,y)}{x_n-x_0}\right |=\left | \frac{\partial f}{\partial x}(\xi_n,y)\right|\leq M$$ where $\xi_n$ is between $x_n$ and $x_0$, and $M=const$ is some upper bound for $\frac{\partial f}{\partial x}(x,y)$ in $[0,1]\times[0,1]$. Here $M$ is summable because $[0,1]$ has finite measure. Now we apply LDC theorem to the sequence $$g_n(y)=\frac{f(x_n,y)-f(x_0,y)}{x_n-x_0}\to \frac{\partial f}{\partial x}(x_0,y)$$ and take the limit under the integral sign: $$\left(\frac{d}{dx}\int\limits_{0}^{1}{f(x,y)dy}\right)_{|x=x_0}=\lim\limits_{n\to\infty}{\int\limits_{0}^{1}{g_n(y)dy}}=\int\limits_{0}^{1}{\lim\limits_{n\to\infty}{g_n(y)}dy}=\int\limits_{0}^{1}{\frac{\partial f}{\partial x}(x_0,y)dy}$$ Finally, since $x_0\in(0,1)$ was arbitrary, the result follows.

Answer 2

Measurability of $\partial_x f(x, \cdot)$ is correct. So I will only say something about $\frac{d}{dx} \int_0^1 f(x,\cdot) = \int_0^1 \frac\partial{\partial x} f(x, \cdot)$.

Your approach is correct, just little bit incomplete. You can't use a sequence as dominating "function" if its $L_1$ norm is unbounded.

Assume $\frac\partial{\partial x} f$ is bounded by $M$. Fix some $x\in[0,1]$. Let $0\ne h_n\to 0$ and define $f_n(y) = \frac{f(x+ h_n, y) - f(x,y)}{h_n}$ for $n\in\mathbb N$ and $y\in[0,1]$. By mean value theorem we have $$ |f_n(y)| = \left|\frac\partial{\partial x} f(\xi_n(y), y) \right| \le M $$ for a suitable $\xi_n(y)$ in the open interval span by $x$ and $x+h_n$ which depends on $n$ and $y$. That's $f_n$ is dominated by $M$. Since the integral domain has finite measure, the constant function $M$ is integrable and the dominated convergence theorem applies: \begin{align} \lim_{n\to\infty} \frac{1}{h_n} \left( \int_0^1 f(x+h_n, \cdot) - \int_0^1 f(x, \cdot) \right) &= \lim_{n\to\infty} \int_0^1 f_n \\ &= \int_0^1 \lim_{n\to\infty} f_n = \int_0^1 \frac\partial{\partial x} f(x, \cdot). \end{align} Since $h_n$ was arbitrary, it follows that $x\mapsto \int_0^1 f(x, \cdot)$ is differentiable at $x$ with $$ \frac{d}{dx} \int_0^1 f(x, \cdot) = \int_0^1 \frac\partial{\partial x} f(x, \cdot). $$