# Define $g(y)=\int_E f(x+y)\ dx$ where $f:\Bbb{R}^n\to\Bbb{R}$ is Lebesgue integrable on compact sets and $E$ is compact with positive measure.

The whole question is -

Let $$f:\Bbb{R}^n\to\Bbb{R}$$ be measurable function and $$E$$ be a compact set of positive Lebesgue measure. Define $$g(y)=\int\limits_E f(x+y)\ dx$$ Prove that $$g$$ is continuous if $$f$$ is continuous. If $$f$$ is Lebesgue integrable on compact sets (instead of being continuous), then is $$g$$ still continuous?

I have proved the first part using DCT and the fact that measure of $$E$$ is finite (since $$E$$ is compact). But I'm stuck with the second part. I was trying with some examples like $$f(x)=\|x\|^{-1/2}$$ but I'm getting $$g$$ to be continuous.

Can anyone help with an idea how to disprove or Prove the second part? Thanks for help in advance.

• Hint: You can approximate $f$ on compact sets (in $L^{1}$ norm ) by a smooth function. Uniform limits of continuous functions are continuous so the answer is YES. Commented Apr 15, 2022 at 10:02
• Thanks for the hint Commented Apr 15, 2022 at 10:40

If $$f$$ is continuous and a sequence $$(y_n)_{n\in\mathbb{N}}$$ that converges to $$y$$, then \begin{align*}\newcommand{\dd}{\mathrm{d}} |g(y) - g(y_{n})| = \Big|\int_{E} f(x+y) - f(x+y_{n}) \,\dd{x}\Big| \leq \int_{E}| f(x+y) - f(x+y_{n})| \,\dd{x} \end{align*} Pointwise we have $$f(x+y_n) \to f(x+y)$$. Hence, by Lesbesgue's theorem \begin{align*} |g(y) - g(y_{n})| \leq \int_{E}| f(x+y) - f(x+y_{n})| \,\dd{x} \to 0, \end{align*} which implies the continuity of $$g$$.

For your updated question note that is sufficient to show that $$g$$ is continuous in $$0$$. Moreover, we have the inequality $$|g(0) - g(y)| \leq \|f - f(\cdot + y)\|_{L^{1}}.$$ Hence if we can show that $$\|f - f(\cdot + y)\|_{L^{1}}$$ converges to $$0$$ for $$y\to 0$$, then the original assertions follows.

You have to start that the new assertion is true for indicator functions, which implies that it is also true for step functions. It is sufficient to prove this for $$n$$-dimensional rectangles.

Note that every $$f \in L^1(\mathbb{R}^n)$$ can be approximated by step functions. Let $$(s_n)_{n\in\mathbb{N}}$$ be such a sequence that converges to $$f$$. For given $$\epsilon >0$$ we choose $$n\in\mathbb{N}$$ large enough such that $$\|f - s_n\|_{L^1} < \epsilon$$ and choose $$y$$ small enough ($$\|y\| < \delta$$) such that $$\|s_n - s_n(\cdot + y)\| < \epsilon$$. Moreover, note that $$\|s_n(\cdot + y) - f(\cdot + y)\|_{L^1} = \|s_n - f\|_{L^1}$$. Hence, \begin{align*} \|f - f(\cdot + y)\|_{L^1} &= \|f - s_n + s_n - s_n(\cdot + y) + s_n(\cdot + y) - f(\cdot + y)\|_{L^1} \\ &\leq \underbrace{\|f - s_n\|_{L^1}}_{< \epsilon} + \underbrace{\|s_n - s_n(\cdot + y)\|_{L^1}}_{< \epsilon} + \underbrace{\|s_n(\cdot + y) - f(\cdot + y)\|_{L^1}}_{<\epsilon} \\ &\leq 3\epsilon, \end{align*} which proves the assertion.

Following the hint of @Kavi Rama Murthy Sir's comment, I'm writting the solution (of the second part)-

Proof of the first part:

$$f$$ is given to be continuous. Let $$y_0$$ be fixed and $$\{y_m\}$$ be asequence converging to $$y_0$$. We've to show $$g(y_m)\to g(y_0)$$

Let $$C$$ be a compact neighbourhood of $$y_0$$, then there is $$n_0\in\Bbb{N}$$ such that $$y_m\in C\ \forall m\ge n_0$$. WLOG assume $$\{y_m\}\subset C$$.

$$f$$ is uniformly continuous on the compact set $$E+C$$. Choose $$\epsilon>0$$, there exists $$\delta>0$$ such that $$|f(z)-f(w)|<\epsilon\ \forall z,w\in E+C$$ with $$\|z-w\|<\delta$$. Now for that $$\delta>0$$ there exists $$m_0\in\Bbb{N}$$ such that $$|y_m-y_0|<\delta\ \forall m\ge m_0$$.

So using the above observations, we have $$\forall x\in E$$, $$|f(x+y_m)-f(x+y_0)|<\epsilon\ \forall m\ge m_0$$.

Hence we have $$|g(y_m)-g(y_0)|\le \int\limits_{E} |f(x+y_m)-f(x+y_0)|\ dx\le \epsilon\lambda(E)\ \forall m\ge m_0$$

Proof of the second part:

Let $$f$$ be Lebesgue integrable on compact sets i.e. $$f\in L^1(C)$$ for all compact sets $$C$$.

Let us fix $$y_0$$, to show $$g$$ is continuous at $$y_0$$. Let $$C$$ be a compact neighbourhood of $$y_0$$. Then $$E+C$$ is compact. Then $$f\in L^1(E+C)\implies \exists$$ a sequence of continuous function $$\{f_m\}$$ such that $$\|f_m-f\|_{L^1(E+C)}\to0$$ (since $$C(E+C)$$ is dense in $$L^1(E+C)$$).

Choose $$\epsilon>0\ \exists N_0$$ such that $$\|f_m-f\|_{L^1(E+C)}<\epsilon\ \forall m\ge N_0$$

Define $$g_m(y)=\int\limits_Ef_m(x+y)\ dx\ \forall y\in C$$ Then $$g_m$$ is continuous on $$C$$ (by first part). We claim that $$g_m$$ converges to $$g$$ uniformly on $$C$$.

Now forall $$y\in C$$, $$|g_m(y)-g(y)|\le \|f_m-f\|_{L^1(E+y)}\le \|f_m-f\|_{L^1(E+C)}<\epsilon\ \forall m\ge N_0$$. This implies $$g_m$$ converges to $$g$$ uniformly on $$C$$. Hence, $$g$$ is continuous on $$C$$, in particular at $$y_0$$.