[proofplan] Uniform convergence first gives continuity of the limit by the standard $\varepsilon/3$ argument: approximate $f$ by one continuous function $f_N$, use continuity of $f_N$ at the point, and then return to $f$. Once $f$ is continuous, all functions involved are bounded Borel functions on the compact interval $[a,b]$, hence Lebesgue integrable. Finally, the integral error is controlled by the uniform error times the length of the interval: \begin{align*} \left|\int_a^b f_n\,d\mathcal{L}^1-\int_a^b f\,d\mathcal{L}^1\right|\le (b-a)\sup_{x\in[a,b]}|f_n(x)-f(x)|. \end{align*} [/proofplan]

[step:Prove that the uniform limit is continuous] Fix $x_0\in[a,b]$ and let $\varepsilon>0$. Since $f_n\to f$ uniformly on $[a,b]$, there exists $N\in\mathbb{N}$ such that \begin{align*} \sup_{y\in[a,b]}|f_N(y)-f(y)|<\frac{\varepsilon}{3}. \end{align*} Because $f_N:[a,b]\to\mathbb{R}$ is continuous at $x_0$, there exists $\delta>0$ such that, for every $x\in[a,b]$ with $|x-x_0|<\delta$, \begin{align*} |f_N(x)-f_N(x_0)|<\frac{\varepsilon}{3}. \end{align*} For such $x$, the triangle inequality gives \begin{align*} |f(x)-f(x_0)|\le |f(x)-f_N(x)|+|f_N(x)-f_N(x_0)|+|f_N(x_0)-f(x_0)|. \end{align*} Each of the first and third terms is bounded by the uniform estimate, and the middle term is bounded by continuity of $f_N$ at $x_0$. Hence \begin{align*} |f(x)-f(x_0)|<\varepsilon. \end{align*} Thus $f$ is continuous at $x_0$. Since $x_0\in[a,b]$ was arbitrary, $f:[a,b]\to\mathbb{R}$ is continuous. [/step]

[step:Verify that all integrals are well defined] For every $n\in\mathbb{N}$, the function $f_n:[a,b]\to\mathbb{R}$ is continuous, hence Borel measurable and bounded on the compact interval $[a,b]$. The previous step shows that $f:[a,b]\to\mathbb{R}$ is also continuous, hence Borel measurable and bounded on $[a,b]$. Since \begin{align*} \mathcal{L}^1([a,b])=b-a<\infty, \end{align*} each bounded Borel function $f_n$ and $f$ is Lebesgue integrable over $[a,b]$ with respect to $\mathcal{L}^1$. [/step]

[step:Bound the integral error by the uniform error]For each $n\in\mathbb{N}$, define \begin{align*} s_n:=\sup_{x\in[a,b]}|f_n(x)-f(x)|. \end{align*} By uniform convergence, $s_n\to 0$ as $n\to\infty$. Using linearity of the Lebesgue integral and the absolute value estimate for integrals, we obtain \begin{align*} \left|\int_a^b f_n(x)\,d\mathcal{L}^1(x)-\int_a^b f(x)\,d\mathcal{L}^1(x)\right|=\left|\int_a^b (f_n(x)-f(x))\,d\mathcal{L}^1(x)\right|. \end{align*} Therefore \begin{align*} \left|\int_a^b (f_n(x)-f(x))\,d\mathcal{L}^1(x)\right|\le \int_a^b |f_n(x)-f(x)|\,d\mathcal{L}^1(x). \end{align*} For every $x\in[a,b]$, the definition of $s_n$ gives $|f_n(x)-f(x)|\le s_n$. Monotonicity of the Lebesgue integral then gives \begin{align*} \int_a^b |f_n(x)-f(x)|\,d\mathcal{L}^1(x)\le \int_a^b s_n\,d\mathcal{L}^1(x). \end{align*} Since $s_n$ is constant as a function of $x$ and $\mathcal{L}^1([a,b])=b-a$, \begin{align*} \int_a^b s_n\,d\mathcal{L}^1(x)=s_n(b-a). \end{align*} Combining these estimates, \begin{align*} \left|\int_a^b f_n(x)\,d\mathcal{L}^1(x)-\int_a^b f(x)\,d\mathcal{L}^1(x)\right|\le (b-a)s_n. \end{align*} Since $s_n\to 0$ and $b-a$ is fixed, the right-hand side tends to $0$.[/step]

[guided]The purpose of this step is to convert uniform convergence into a numerical bound on the difference of the integrals. For each $n\in\mathbb{N}$, define the uniform error \begin{align*} s_n:=\sup_{x\in[a,b]}|f_n(x)-f(x)|. \end{align*} The hypothesis $f_n\to f$ uniformly on $[a,b]$ means exactly that $s_n\to 0$ as $n\to\infty$. We now estimate the integral difference. By linearity of the Lebesgue integral applied to the integrable functions $f_n$ and $f$, \begin{align*} \int_a^b f_n(x)\,d\mathcal{L}^1(x)-\int_a^b f(x)\,d\mathcal{L}^1(x)=\int_a^b (f_n(x)-f(x))\,d\mathcal{L}^1(x). \end{align*} Taking absolute values and applying the integral absolute value estimate gives \begin{align*} \left|\int_a^b (f_n(x)-f(x))\,d\mathcal{L}^1(x)\right|\le \int_a^b |f_n(x)-f(x)|\,d\mathcal{L}^1(x). \end{align*} The uniform error $s_n$ controls the integrand at every point of the interval: for every $x\in[a,b]$, \begin{align*} |f_n(x)-f(x)|\le s_n. \end{align*} By monotonicity of the Lebesgue integral, \begin{align*} \int_a^b |f_n(x)-f(x)|\,d\mathcal{L}^1(x)\le \int_a^b s_n\,d\mathcal{L}^1(x). \end{align*} Because $s_n$ is a constant with respect to $x$, and because the one-dimensional Lebesgue measure of $[a,b]$ is $b-a$, we have \begin{align*} \int_a^b s_n\,d\mathcal{L}^1(x)=s_n\mathcal{L}^1([a,b])=s_n(b-a). \end{align*} Combining the preceding estimates yields \begin{align*} \left|\int_a^b f_n(x)\,d\mathcal{L}^1(x)-\int_a^b f(x)\,d\mathcal{L}^1(x)\right|\le (b-a)s_n. \end{align*} Since $s_n\to 0$ and $b-a$ is a fixed positive real number, $(b-a)s_n\to 0$. Therefore the absolute difference between the two integrals tends to $0$.[/guided]

[step:Conclude convergence of the integrals] The estimate from the previous step shows that \begin{align*} 0\le \left|\int_a^b f_n(x)\,d\mathcal{L}^1(x)-\int_a^b f(x)\,d\mathcal{L}^1(x)\right|\le (b-a)s_n. \end{align*} Since $(b-a)s_n\to 0$, the squeeze principle gives \begin{align*} \lim_{n\to\infty}\left|\int_a^b f_n(x)\,d\mathcal{L}^1(x)-\int_a^b f(x)\,d\mathcal{L}^1(x)\right|=0. \end{align*} Equivalently, \begin{align*} \lim_{n\to\infty}\int_a^b f_n(x)\,d\mathcal{L}^1(x)=\int_a^b f(x)\,d\mathcal{L}^1(x). \end{align*} Together with the continuity of $f$ proved above, this proves the theorem. [/step]